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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 $2$, $1011_2 \otimes 1011_2 = 1000101_2$:
$$
\begin{aligned}
\phantom{\otimes 1111} 1011_2 \\
\otimes \phantom{1111} 1011_2 \\
\hline
\phantom{\otimes 1111} 1011_2 \\
\phantom{\otimes 111} 1011_2 \phantom{9} \\
\oplus \phantom{1} 1011_2 \phantom{999} \\
\hline
\phantom{\otimes 11} 1000101_2 \\
\end{aligned}
$$
Further we define $P(n) = 11^{\otimes n} = \overbrace{11\otimes 11\otimes \ldots \otimes 11}^n$. 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$1011_2 \otimes 1011_2 = 1000101_2$：
$$
\begin{aligned}
\phantom{\otimes 1111} 1011_2 \\
\otimes \phantom{1111} 1011_2 \\
\hline
\phantom{\otimes 1111} 1011_2 \\
\phantom{\otimes 111} 1011_2 \phantom{9} \\
\oplus \phantom{1} 1011_2 \phantom{999} \\
\hline
\phantom{\otimes 11} 1000101_2 \\
\end{aligned}
$$
进一步定义$P(n) = 11^{\otimes n} = \overbrace{11\otimes 11\otimes \ldots \otimes 11}^n$，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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