codeforces 913 A Modular Exponentiation

Description

The following problem is well-known: given integers n and m, calculate



where 2n = 2⋅2⋅...⋅2 (n factors), and denotes the remainder of division of x by y.

You are asked to solve the “reverse” problem. Given integers n and m, calculate



Input

The first line contains a single integer n (1 ≤ n ≤ 108).

The second line contains a single integer m (1 ≤ m ≤ 108).

Output

Output a single integer — the value of

.

Examples

Input
4 42
Output
10

Input
1 58
Output
0

Input
98765432 23456789
Output
23456789

Note

In the first example, the remainder of division of 42 by 24 = 16 is equal to 10.

In the second example, 58 is divisible by 21 = 2 without remainder, and the answer is 0.

题意：输入n，m求m mod 2n

思路：因为m最大只有108，所以对于很大的n大可不必将2n计算出来，只用判断一下是否比m大就行了。

#include <stdio.h>

int main() {
int n, m;
//m mod 2^n
scanf("%d %d", &n, &m);
int n2 = 1;
for (int i = 0; i < n; ++i) {//做n次2的幂运算
if (n2 * 2 <= m)n2=n2 * 2;
else { n2 = n2 * 2; break; }//当n2大于m时中断计算
}
printf("%d", m%n2);
}