Period of permutation

Let $i_1,i_2,i_3,\dots ,i_n$ be a permutation of the set $\{1,2,3,...,n\}$ where $n$ is a positive integer. A permutation can also be viewed as a function $f$. We denote this permutation by the 2-by-$n$ array $$f=\left(\begin{array}{cccc}1& 2& \cdots & n\\ i_1& i_2& \cdots & i_n\end{array}\right)$$ It maps number 1 to $i_1$, number 2 to $i_2$ and so on. When this function operates on a permutation of set $1,2,3,...,n$, that is $x_1,x_2,x_3,\cdots ,x_n$, a new permutation $y_1,y_2,y_3,\cdots ,y_n$ can be obtained. Here $y_1=f(x_1),y_2=f(x_2),\cdots ,y_n=f(x_n)$.

For example, when $n=4$ and there is a permutation $2,4,1,3$. We make this permutation operate on a permutation $1,2,3,4$ continuously, and it will be $$1,2,3,4\to 2,4,1,3\to 4,3,2,1\to 3,1,4,2\to 1,2,3,4\to \cdots$$ Obviously after 4 operations the result permutation returns to the initial one.

Let $F(l,p)$ be the number of permutations of the set $\{1,2,3,...,l\}$ which makes the permutation $1,2,3,\cdots ,l$ returns to itself after $p$ operations. For instance, $F(4,3)=9$, as there are 9 permutations meet the condition: $(1,2,3,4)$, $(1,3,4,2)$, $(1,4,2,3)$, $(2,3,1,4)$, $(2,4,3,1)$, $(3,1,2,4)$, $(3,2,4,1)$, $(4,1,3,2)$, $(4,2,1,3)$. You are also given $F(5,5)=25$, $F(10,6)=625176$, $F(20,11)=609493248001$.

Let $S(L,P)={\sum }_{l=1}^L{\sum }_{p=1}^{P}F(l,p)$. Find $S(10000,1000)mod1000000007$.

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Thanks to baihacker for the idea.