A permutation $\pi$ of $\{1,\dots ,n\}$ can be represented in one-line notation as $\pi (1),\dots ,\pi (n)$. If all $n!$ permutations are written in lexicographic order then $\text{rank}(\pi )$ is the position of $\pi$ in this 1-based list.

For example, $\text{rank}(2,1,3)=3$ because the six permutations of $\{1,2,3\}$ in lexicographic order are: $$1,2,31,3,22,1,32,3,13,1,23,2,1$$

For a positive integer $m$, we define the following permutation of $\{1,\dots ,n\}$ with $n=\frac{m(m+1)}{2}$: $$\begin{array}{rl}\sigma (i)& =\{\begin{array}{ll}\frac{k(k-1)}{2}+1& \text{if }i=\frac{k(k+1)}{2}\text{ for }k\in \{1,\dots ,m\};\\ i+1& \text{otherwise};\end{array}\\ \tau (i)& =(({10}^9+7)imodn)+1\\ \pi (i)& ={\tau }^{-1}(\sigma (\tau (i)))\end{array}$$ where ${\tau }^{-1}$ is the inverse permutation of $\tau$.

Define ${\displaystyle P(m)=\sum _{k=1}^{m!}\text{rank}({\pi }^k)}$, where ${\pi }^k$ is the permutation arising from applying $\pi$ $k$ times.
For example, $P(2)=4$, $P(3)=780$ and $P(4)=38810300$.

Find $P(100)$. Give your answer modulo $({10}^9+7)$.