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Problem 813

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XOR-Powers

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 ${10}^9+7$.

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异或幂

记$x\oplus y$为$x$和$y$按位异或的结果。

我们定义一种新运算，称为$x$和$y$的异或积并记作$x\otimes y$。这种运算类似于对$x$和$y$的二进制表示做长乘法，但是将其中的相加替换为异或。

例如，$11\otimes 11=69$，或用二进制表示写作${1011}_2\otimes {1011}_2={1000101}_2$：
$$\begin{array}{r}\phantom{\otimes 1111}{1011}_2\\ \otimes \phantom{1111}{1011}_2\\ \phantom{\otimes 1111}{1011}_2\\ \phantom{\otimes 111}{1011}_2\phantom{9}\\ \oplus \phantom{1}{1011}_2\phantom{999}\\ \phantom{\otimes 11}{1000101}_2\end{array}$$
进一步定义$P(n)={11}^{\otimes n}=\stackrel{n}{\overbrace{11\otimes 11\otimes \dots \otimes 11}}$，例如$P(2)=69$。

求$P(8^{12}\cdot {12}^8)$，并将你的答案对${10}^9+7$取余。

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