We use $x\oplus y$ to be the bitwise XOR of $x$ and $y$.

Define the XOR-product of $x$ and $y$, denoted by $x\otimes y$, similar to a long multiplication in base $2$, except that the intermediate results are XORed instead of the usual integer addition.

For example, $11\otimes 11=69$, or in base $2$, ${1011}_2\otimes {1011}_2={1000101}_2$:

$$\begin{array}{r}\phantom{\otimes 1111}{1011}_2\\ \otimes \phantom{1111}{1011}_2\\ \phantom{\otimes 1111}{1011}_2\\ \phantom{\otimes 111}{1011}_2\phantom{9}\\ \oplus \phantom{1}{1011}_2\phantom{999}\\ \phantom{\otimes 11}{1000101}_2\end{array}$$ Further we define $P(n)={11}^{\otimes n}=\stackrel{n}{\overbrace{11\otimes 11\otimes \dots \otimes 11}}$. For example $P(2)=69$.

Find $P(8^{12}\cdot {12}^8)$. Give your answer modulo ${10}^9+7$.