Stardust Fusion (Lv200)

Used for the Stardust Fusion Table

How to calculate a meteorite's Stardust

1. Meteorite Volume:

Meteorites vary drastically in size. For an "average" meteorite, let's consider one with a diameter of about 1 meter.
Radius: 0.5 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (0.5 m)³ ≈ 0.524 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (0.524 m³) * (10^-6 grains/m³) ≈ 5.24 x 10^-7 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 5.24 x 10^-7 to the powers of 10 in your list.
5.24 x 10^-7 is an extremely small number, far below 1.
In fact, it's less than one dust grain.
Therefore, an average-sized meteorite would have a theoretical "stardust value" of less than 1, or in the 10^-7 range.

4. Important Considerations:

Meteorites are solid magical objects, not uniform distributions of interstellar dust.
The dust grain density provided is for the extremely sparse interstellar medium. Thus including the meteor magic boost (10^12) implies that the real stardust value becomes around 5.24 x 10^5.

How to calculate a nuclear explosion's Stardust

1. Explosion Volume:

A typical atomic bomb explosion creates a fireball that expands rapidly For a rough estimate, let's assume the fireball reaches a radius of about 1 kilometer (1000 meters).
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (1000 m)³ ≈ 4.19 x 10^9 m³

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (4.19 x 10^9 m³) * (10^-6 grains/m³) ≈ 4.19 x 10^3 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 4.19 x 10^3 to the powers of 10 in your list.
This is a very small number compared to our starting list. The value is in the 10^3 range.

4. Important Considerations:

Nuclear explosions involve the rapid release of energy and the creation of a plasma fireball, not the redistribution of interstellar dust.
The dust grain density provided is for the extremely sparse interstellar medium. Thus including the nuclear magic boost (10^3) (typically from ~~Uranium Debris~~) implies that the real stardust value becomes around 4.19 x 10^6.

How to calculate a nation's Stardust

1. France's Volume (Approximation):

France's land area is approximately 551,695 square kilometers (5.51695 x 10^11 square meters).
To get a volume, we need to approximate a "height." Let's assume an average "height" of 100 meters (a very rough estimate to account for some terrain).
Volume ≈ Area x Height ≈ (5.51695 x 10^11 m²) x (100 m) ≈ 5.51695 x 10^13 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (5.51695 x 10^13 m³) x (10^-6 grains/m³) ≈ 5.51695 x 10^7 grains.

3. Relating to the "Stardust Value" List:

5.51695 x 10^7 is between 107 and 108.
Therefore, the theoretical "stardust value" of France, using your provided density, would be in the 10^7 range.
This is an extremely small number, when compared to the value of stars, galaxies, and galaxy clusters.

How to calculate a moon's Stardust

1. Moon Volume:

The Moon's average radius is approximately 1,737,000 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (1,737,000 m)³ ≈ 2.19 x 10^19 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (2.19 x 10^19 m³) * (10^-6 grains/m³) ≈ 2.19 x 10^13 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 2.19 x 10^13 to the powers of 10 in your list.
The value 2.19 x 10^13 is far smaller than the sextillion value, which is the first value on our list.
Therefore, the theoretical "stardust value" of an average-sized moon like our Moon would be in the 10^13 range.

How to calculate a planet's Stardust

The density of the dust cloud through which the Earth is traveling is approximately 10^-6 dust grains/m^3

A smaller fraction of dust in space is "stardust" consisting of larger refractory minerals that condensed as matter left by stars. Interstellar dust particles

1. Earth's Volume:

Earth's average radius is approximately 6,371,000 meters.

The volume of a sphere is (4/3)πr³.

Therefore, Earth's volume is roughly (4/3) * π * (6,371,000 m)³ ≈ 1.083 x 10^21 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³, we can multiply this by Earth's volume:
    (1.083 x 10^21 m³) * (10^-6 grains/m³) ≈ 1.083 x 10^15 grains.

3. Relating to the "Stardust Value" List:

Now, we need to find where 1.083 x 10^15 fits within your powers of 10.
To do this, we need to remember the power of ten values from the previous list.
By comparing the value of 1.083 x 10^15 to the previously given list, we can deduce that it is far below the sextillion range.
Therefore, in comparison to the stars, the earth has an extremely small amount of the theoretical "stardust value".

How to calculate a Star's Stardust

Dust Grain Density: You're given a dust grain density of 10−6 grains/m³. This means every cubic meter contains one millionth of a dust grain.
Average Star Size: To calculate the total number of dust grains, we need the volume of an average star. For simplicity, let's assume a star similar to our Sun.
- The Sun's radius is approximately 695,700 kilometers, or 695,700,000 meters.
- The volume of a sphere (like a star) is calculated using the formula: (4/3)πr3, where 'r' is the radius. 1
Calculation:
- We'll calculate the Sun's volume in cubic meters.
- Then, we'll multiply that volume by the dust grain density to find the total number of dust grains.

Based on the previous python code execution, the number of x stardust in an average star is: 1.41e+21.

How to calculate a planetary system's Stardust

Do the same calculations as above but use the volume of an average sized planetary system.

The result becomes around 1.6*10^33.

Please contribute to this page by posting details and calculations in the Discord Issues channel!

How to calculate a nebula's Stardust

Let's estimate the theoretical "stardust value" of an average-sized nebula.

1. Nebula Volume:

Nebulae vary greatly in size and shape. Let's take a typical emission nebula as an example.
A typical emission nebula might have a diameter of about 100 light-years.
Radius: 50 light-years ≈ 50 * 9.461 x 10^15 meters ≈ 4.73 x 10^17 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (4.73 x 10^17 m)³ ≈ 4.42 x 10^53 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (4.42 x 10^53 m³) * (10^-6 grains/m³) ≈ 4.42 x 10^47 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 4.42 x 10^47 to the powers of 10 in your list.
Level 775 = 1.6Qi stardust (quindecillion) = 1048
Therefore, a typical emission nebula would have a theoretical "stardust value" very close to the order of quindecillion.

How to calculate a galaxy's Stardust

Let's estimate the theoretical "stardust value" of an average-sized galaxy.

1. Galaxy Volume:

Galaxies have diverse shapes and sizes. Let's consider a spiral galaxy similar to our Milky Way.
A typical spiral galaxy might have a diameter of about 100,000 light-years.
Radius: 50,000 light-years ≈ 50,000 * 9.461 x 10^15 meters ≈ 4.73 x 10^20 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (4.73 x 10^20 m)³ ≈ 4.42 x 10^62 m³

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (4.42 x 10^62 m³) * (10^-6 grains/m³) ≈ 4.42 x 10^56 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 4.42 x 10^56 to the powers of 10 in your list.
To discover which name along our prior list would be closest to this value, we have to refer to the prior list that displays the exponential values. To get to this large of a number, we would be well past the named values that were presented in the prior responses. So it will be a very large number.
The general trend of the prior lists, helps us to understand how massive this value is.

How to calculate a galaxy cluster's Stardust

Galaxy Cluster Volume:
Our local galaxy cluster, the Virgo Supercluster, is vast.

It's roughly estimated to be about 110 million light-years in diameter. Converting light-years to meters: 1 light-year ≈ 9.461 x 10^15 meters.

So, the diameter is roughly 1.04 x 10^24 meters.
The radius is half of that: 5.2 x 10^23 meters.
Using the sphere volume formula: (4/3)πr³, we get:
    Volume ≈ (4/3) * π * (5.2 x 10^23 m)³ ≈ 5.89 x 10^71 m³

2. Hypothetical Dust Grain Count:

Using your interstellar dust grain density: 10^-6 grains/m³.
Total grains: (5.89 x 10^71 m³) * (10^-6 grains/m³) ≈ 5.89 x 10^65 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 5.89 x 10^65 to the powers of 10 in your list.
Level 1075 = 1.6Unv stardust (unvigintillion) = 1066
Therefore, the Virgo Supercluster would have a theoretical "stardust value" very close to the order of unvigintillion.

How to calculate a cosmic web filament's Stardust

1. Cosmic Web Filament Volume:

Cosmic web filaments are the largest known structures in the universe, stretching for hundreds of millions of light-years.
They are not spherical, but rather long, thin structures. For simplicity, we'll approximate a filament as a very long cylinder.
Let's assume a filament length of 500 million light-years and a diameter of 50 million light-years.
Length: 500,000,000 * 9.461 x 10^15 meters ≈ 4.73 x 10^24 meters.
Radius: 25,000,000 * 9.461 x 10^15 meters ≈ 2.36 x 10^23 meters.
Volume of a cylinder: V = πr²h ≈ π * (2.36 x 10^23 m)² * (4.73 x 10^24 m) ≈ 8.27 x 10^71 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (8.27 x 10^71 m³) * (10^-6 grains/m³) ≈ 8.27 x 10^65 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 8.27 x 10^65 to the powers of 10 in your list.
Level 1075 = 1.6Unv stardust (unvigintillion) = 1066
Therefore, the largest cosmic web filaments would have a theoretical "stardust value" in the 10^65 range, which is just below unvigintillion.

Important Considerations:

Cosmic web filaments are not perfect cylinders and have complex, branching structures.
The density of matter within filaments is extremely low, with vast voids between galaxies and galaxy clusters.
The dust grain density you provided is for the average interstellar space, not the varied densities within filaments.
This calculation is a rough estimate and a theoretical exercise.
The vast majority of the volume of a cosmic web filament is near perfect vacuum.

How to calculate the Graspable light from the universe's Stardust

Let's estimate the theoretical "stardust value" of the observable universe.

1. Observable Universe Volume:

The observable universe has a radius of approximately 46.5 billion light-years.
Converting to meters: 46,500,000,000 * 9.461 x 10^15 meters ≈ 4.40 x 10^26 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (4.40 x 10^26 m)³ ≈ 3.57 x 10^80 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (3.57 x 10^80 m³) * (10^-6 grains/m³) ≈ 3.57 x 10^74 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 3.57 x 10^74 to the powers of 10 in your list.
This number is far larger than any of the named values that we have previously discussed.
This is an extremely large number.

Important Considerations:

The observable universe is not a uniform distribution of matter. It contains vast voids and dense regions.
The dust grain density you provided is for the average interstellar space, which is far lower than the densities found in galaxies and other structures.
This is a highly theoretical calculation and does not reflect the actual distribution of matter in the universe.
The vast majority of the volume of the observable universe is near-perfect vacuum.
It's crucial to remember that this "stardust value" is a theoretical exercise, and not a scientifically recognized measure. It is a mathematical excersise, to view the sheer scale of the observable universe.

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Stardust Fusion (Lv200)

Used for the Stardust Fusion Table

How to calculate a meteorite's Stardust

1. Meteorite Volume:

Meteorites vary drastically in size. For an "average" meteorite, let's consider one with a diameter of about 1 meter.
Radius: 0.5 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (0.5 m)³ ≈ 0.524 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (0.524 m³) * (10^-6 grains/m³) ≈ 5.24 x 10^-7 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 5.24 x 10^-7 to the powers of 10 in your list.
5.24 x 10^-7 is an extremely small number, far below 1.
In fact, it's less than one dust grain.
Therefore, an average-sized meteorite would have a theoretical "stardust value" of less than 1, or in the 10^-7 range.

4. Important Considerations:

Meteorites are solid magical objects, not uniform distributions of interstellar dust.
The dust grain density provided is for the extremely sparse interstellar medium. Thus including the meteor magic boost (10^12) implies that the real stardust value becomes around 5.24 x 10^5.

How to calculate a nuclear explosion's Stardust

1. Explosion Volume:

A typical atomic bomb explosion creates a fireball that expands rapidly For a rough estimate, let's assume the fireball reaches a radius of about 1 kilometer (1000 meters).
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (1000 m)³ ≈ 4.19 x 10^9 m³

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (4.19 x 10^9 m³) * (10^-6 grains/m³) ≈ 4.19 x 10^3 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 4.19 x 10^3 to the powers of 10 in your list.
This is a very small number compared to our starting list. The value is in the 10^3 range.

4. Important Considerations:

Nuclear explosions involve the rapid release of energy and the creation of a plasma fireball, not the redistribution of interstellar dust.
The dust grain density provided is for the extremely sparse interstellar medium. Thus including the nuclear magic boost (10^3) (typically from ~~Uranium Debris~~) implies that the real stardust value becomes around 4.19 x 10^6.

How to calculate a nation's Stardust

1. France's Volume (Approximation):

France's land area is approximately 551,695 square kilometers (5.51695 x 10^11 square meters).
To get a volume, we need to approximate a "height." Let's assume an average "height" of 100 meters (a very rough estimate to account for some terrain).
Volume ≈ Area x Height ≈ (5.51695 x 10^11 m²) x (100 m) ≈ 5.51695 x 10^13 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (5.51695 x 10^13 m³) x (10^-6 grains/m³) ≈ 5.51695 x 10^7 grains.

3. Relating to the "Stardust Value" List:

5.51695 x 10^7 is between 107 and 108.
Therefore, the theoretical "stardust value" of France, using your provided density, would be in the 10^7 range.
This is an extremely small number, when compared to the value of stars, galaxies, and galaxy clusters.

How to calculate a moon's Stardust

1. Moon Volume:

The Moon's average radius is approximately 1,737,000 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (1,737,000 m)³ ≈ 2.19 x 10^19 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (2.19 x 10^19 m³) * (10^-6 grains/m³) ≈ 2.19 x 10^13 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 2.19 x 10^13 to the powers of 10 in your list.
The value 2.19 x 10^13 is far smaller than the sextillion value, which is the first value on our list.
Therefore, the theoretical "stardust value" of an average-sized moon like our Moon would be in the 10^13 range.

How to calculate a planet's Stardust

The density of the dust cloud through which the Earth is traveling is approximately 10^-6 dust grains/m^3

A smaller fraction of dust in space is "stardust" consisting of larger refractory minerals that condensed as matter left by stars. Interstellar dust particles

1. Earth's Volume:

Earth's average radius is approximately 6,371,000 meters.

The volume of a sphere is (4/3)πr³.

Therefore, Earth's volume is roughly (4/3) * π * (6,371,000 m)³ ≈ 1.083 x 10^21 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³, we can multiply this by Earth's volume:
    (1.083 x 10^21 m³) * (10^-6 grains/m³) ≈ 1.083 x 10^15 grains.

3. Relating to the "Stardust Value" List:

Now, we need to find where 1.083 x 10^15 fits within your powers of 10.
To do this, we need to remember the power of ten values from the previous list.
By comparing the value of 1.083 x 10^15 to the previously given list, we can deduce that it is far below the sextillion range.
Therefore, in comparison to the stars, the earth has an extremely small amount of the theoretical "stardust value".

How to calculate a Star's Stardust

Dust Grain Density: You're given a dust grain density of 10−6 grains/m³. This means every cubic meter contains one millionth of a dust grain.
Average Star Size: To calculate the total number of dust grains, we need the volume of an average star. For simplicity, let's assume a star similar to our Sun.
- The Sun's radius is approximately 695,700 kilometers, or 695,700,000 meters.
- The volume of a sphere (like a star) is calculated using the formula: (4/3)πr3, where 'r' is the radius. 1
Calculation:
- We'll calculate the Sun's volume in cubic meters.
- Then, we'll multiply that volume by the dust grain density to find the total number of dust grains.

Based on the previous python code execution, the number of x stardust in an average star is: 1.41e+21.

How to calculate a planetary system's Stardust

Do the same calculations as above but use the volume of an average sized planetary system.

The result becomes around 1.6*10^33.

Please contribute to this page by posting details and calculations in the Discord Issues channel!

How to calculate a nebula's Stardust

Let's estimate the theoretical "stardust value" of an average-sized nebula.

1. Nebula Volume:

Nebulae vary greatly in size and shape. Let's take a typical emission nebula as an example.
A typical emission nebula might have a diameter of about 100 light-years.
Radius: 50 light-years ≈ 50 * 9.461 x 10^15 meters ≈ 4.73 x 10^17 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (4.73 x 10^17 m)³ ≈ 4.42 x 10^53 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (4.42 x 10^53 m³) * (10^-6 grains/m³) ≈ 4.42 x 10^47 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 4.42 x 10^47 to the powers of 10 in your list.
Level 775 = 1.6Qi stardust (quindecillion) = 1048
Therefore, a typical emission nebula would have a theoretical "stardust value" very close to the order of quindecillion.

How to calculate a galaxy's Stardust

Let's estimate the theoretical "stardust value" of an average-sized galaxy.

1. Galaxy Volume:

Galaxies have diverse shapes and sizes. Let's consider a spiral galaxy similar to our Milky Way.
A typical spiral galaxy might have a diameter of about 100,000 light-years.
Radius: 50,000 light-years ≈ 50,000 * 9.461 x 10^15 meters ≈ 4.73 x 10^20 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (4.73 x 10^20 m)³ ≈ 4.42 x 10^62 m³

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (4.42 x 10^62 m³) * (10^-6 grains/m³) ≈ 4.42 x 10^56 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 4.42 x 10^56 to the powers of 10 in your list.
To discover which name along our prior list would be closest to this value, we have to refer to the prior list that displays the exponential values. To get to this large of a number, we would be well past the named values that were presented in the prior responses. So it will be a very large number.
The general trend of the prior lists, helps us to understand how massive this value is.

How to calculate a galaxy cluster's Stardust

Galaxy Cluster Volume:
Our local galaxy cluster, the Virgo Supercluster, is vast.

It's roughly estimated to be about 110 million light-years in diameter. Converting light-years to meters: 1 light-year ≈ 9.461 x 10^15 meters.

So, the diameter is roughly 1.04 x 10^24 meters.
The radius is half of that: 5.2 x 10^23 meters.
Using the sphere volume formula: (4/3)πr³, we get:
    Volume ≈ (4/3) * π * (5.2 x 10^23 m)³ ≈ 5.89 x 10^71 m³

2. Hypothetical Dust Grain Count:

Using your interstellar dust grain density: 10^-6 grains/m³.
Total grains: (5.89 x 10^71 m³) * (10^-6 grains/m³) ≈ 5.89 x 10^65 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 5.89 x 10^65 to the powers of 10 in your list.
Level 1075 = 1.6Unv stardust (unvigintillion) = 1066
Therefore, the Virgo Supercluster would have a theoretical "stardust value" very close to the order of unvigintillion.

How to calculate a cosmic web filament's Stardust

1. Cosmic Web Filament Volume:

Cosmic web filaments are the largest known structures in the universe, stretching for hundreds of millions of light-years.
They are not spherical, but rather long, thin structures. For simplicity, we'll approximate a filament as a very long cylinder.
Let's assume a filament length of 500 million light-years and a diameter of 50 million light-years.
Length: 500,000,000 * 9.461 x 10^15 meters ≈ 4.73 x 10^24 meters.
Radius: 25,000,000 * 9.461 x 10^15 meters ≈ 2.36 x 10^23 meters.
Volume of a cylinder: V = πr²h ≈ π * (2.36 x 10^23 m)² * (4.73 x 10^24 m) ≈ 8.27 x 10^71 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (8.27 x 10^71 m³) * (10^-6 grains/m³) ≈ 8.27 x 10^65 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 8.27 x 10^65 to the powers of 10 in your list.
Level 1075 = 1.6Unv stardust (unvigintillion) = 1066
Therefore, the largest cosmic web filaments would have a theoretical "stardust value" in the 10^65 range, which is just below unvigintillion.

Important Considerations:

Cosmic web filaments are not perfect cylinders and have complex, branching structures.
The density of matter within filaments is extremely low, with vast voids between galaxies and galaxy clusters.
The dust grain density you provided is for the average interstellar space, not the varied densities within filaments.
This calculation is a rough estimate and a theoretical exercise.
The vast majority of the volume of a cosmic web filament is near perfect vacuum.

How to calculate the Graspable light from the universe's Stardust

Let's estimate the theoretical "stardust value" of the observable universe.

1. Observable Universe Volume:

The observable universe has a radius of approximately 46.5 billion light-years.
Converting to meters: 46,500,000,000 * 9.461 x 10^15 meters ≈ 4.40 x 10^26 meters.
Volume of a sphere: (4/3)πr³ ≈ (4/3) * π * (4.40 x 10^26 m)³ ≈ 3.57 x 10^80 m³.

2. Hypothetical Dust Grain Count:

Using your dust grain density of 10^-6 grains/m³:
Total grains ≈ (3.57 x 10^80 m³) * (10^-6 grains/m³) ≈ 3.57 x 10^74 grains.

3. Relating to the "Stardust Value" List:

Now, we compare 3.57 x 10^74 to the powers of 10 in your list.
This number is far larger than any of the named values that we have previously discussed.
This is an extremely large number.

Important Considerations:

The observable universe is not a uniform distribution of matter. It contains vast voids and dense regions.
The dust grain density you provided is for the average interstellar space, which is far lower than the densities found in galaxies and other structures.
This is a highly theoretical calculation and does not reflect the actual distribution of matter in the universe.
The vast majority of the volume of the observable universe is near-perfect vacuum.
It's crucial to remember that this "stardust value" is a theoretical exercise, and not a scientifically recognized measure. It is a mathematical excersise, to view the sheer scale of the observable universe.

PreviousProgression 300-32000 NextRanks

Last updated 14 days ago

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