Hash Function

A hash function $h\_n$ encrypts a number $A$ consisting of $2n$ bits as follows.

Let $A = (a\_{2n-1}a\_{2n-2} \cdots a\_1a\_0)\_2$, where $a\_i$ is the $i$-th bit of $A$.

The number $B = (b\_{2n-1}b\_{2n-2} \cdots b\_1b\_0)\_2$, also consisting of $2n$ bits, is computed as follows:

$b\_i = a\_i \oplus a\_{2i+1},\text{ for } 0 \le i < n,$ $b\_i = a\_i \oplus a\_{4n-2i-2},\text{ for } n \le i < 2n,$

where $\oplus$ is bitwise exclusive OR (XOR). In other words, $B = A \oplus (a\_0a\_2 \cdots a\_{2n-4}a\_{2n-2}a\_{2n-1}a\_{2n-3} \cdots a\_3a\_1)\_2.$

Next, the number $C = B \oplus \text{RSH}(B)$ is computed, also consisting of $2n$ bits, where $\text{RSH}(B)$ is a cyclic right shift of $B$ by 1 bit. In other words, $C = B \oplus (b\_0b\_{2n-1}b\_{2n-2} \cdots b\_2b\_1)\_2$.

Finally, the hash value is $h\_n(A) = 239A + 153C \bmod (2^{2n-1}-1)$.

For example, let $n=4$ and $A = 00001101\_2 = 13$. Then $B = 00001101\_2 \oplus 11000010\_2 = 11001111\_2 = 207$. Next, $C = 11001111\_2 \oplus 11100111\_2 = 00101000\_2 = 40$. Finally, $h\_4(A) = 239 \times 13 + 153 \times 40 \bmod (2^7-1) = 9227 \bmod 127 = 83$.

Your goal is to invert this hash function. For given $n$ and $H$, find $A$ such that $h\_n(A)=H$.

You are given two integers $n$ and $H$ ($2 \le n \le 16$, $0 \le H < 2^{2n-1}-1$).

It is guaranteed that there exists $A$ ($0 \le A < 2^{2n}$) such that $h\_n(A)=H$.

Print one integer $A$ ($0 \le A < 2^{2n}$) such that $h\_n(A)=H$.

If there are several such $A$, output any of them.