Consider writing a natural number as product of powers of natural numbers with given exponents, additionally requiring different base numbers for each power.

For example, $256$ can be written as a product of a square and a fourth power in three ways such that the base numbers are different.
That is, $256=1^2\times 4^4=4^2\times 2^4={16}^2\times 1^4$

Though $4^2$ and $2^4$ are both equal, we are concerned only about the base numbers in this problem. Note that permutations are not considered distinct, for example ${16}^2\times 1^4$ and $1^4\times {16}^2$ are considered to be the same.

Similarly, $10!$ can be written as a product of one natural number, two squares and three cubes in two ways ($10!=42\times 5^2\times 4^2\times 3^3\times 2^3\times 1^3=21\times 5^2\times 2^2\times 4^3\times 3^3\times 1^3$) whereas $20!$ can be given the same representation in $41680$ ways.

Let $F(n)$ denote the number of ways in which $n$ can be written as a product of one natural number, two squares, three cubes and four fourth powers.

You are given that $F(25!)=4933$, $F(100!)mod1000000007=693952493$,
and $F(1000!)mod1000000007=6364496$.

Find $F(1000000!)mod1000000007$.