
Equiproduct Partition
Let $\theta=\sqrt{-2}$.
Define $T$ to be the set of numbers of the form $a+b\theta$, where $a$ and $b$ are integers and either $a\gt 0$, or $a=0$ and $b\gt 0$. For a set $S \subseteq T$ and element $z \in T$, define $p(S,z)$ to be the number of ways of choosing two distinct elements from $S$ with product either $z$ or $-z$.
For example if $S=\{1,2,4\}$ and $z=4$, there is only one valid pair of elements with product $\pm4$, namely $1$ and $4$. Thus, in this case $p(S,z)=1$.
For another example, if $S=\{1,\theta,1+\theta,2-\theta\}$ and $z=2-\theta$, we have $1\cdot(2-\theta)=z$ and $\theta\cdot(1+\theta)=-z$, giving $p(S,z)=2$.
Let $A$ and $B$ be two sets satisfying the following conditions:
- $1 \in A$
- $A \cap B = \emptyset$
- $A \cup B = T$
- $p(A,z) = p(B,z)$ for all $z\in T$
Remarkably, these four conditions uniquely determine the sets $A$ and $B$.
Let $F_n$ be the set of the first $n$ factorials: $F_n=\{1!,2!,\dots,n!\}$, and define $G(n)$ to be the sum of all elements of $F_n\cap A$.
You are given $G(4) = 25$, $G(7) = 745$, and $G(100) \equiv 709772949 \pmod{10^9+7}$.
Find $G(10^8)$ and give your answer modulo $10^9+7$.
