Define $g(n, m)$ to be the largest integer $k$ such that $2^k$ divides $\binom{n}m$. For example, $\binom{12}5 = 792 = 2^3 \cdot 3^2 \cdot 11$, hence $g(12, 5) = 3$. Then define $F(n) = \max \{ g(n, m) : 0 \le m \le n \}$. $F(10) = 3$ and $F(100) = 6$.

Let $S(N)$ = $\displaystyle\sum_{n=1}^N{F(n)}$. You are given that $S(100) = 389$ and $S(10^7) = 203222840$.

Find $S(10^{N^3}) \bmod 1000000007$.