A. Modular Exponentiation

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

解题说明：水题，按照题目意思求解即可。

#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
using namespace std;

int main()

{
int n, m, x;
scanf("%d %d", &n, &m);
x = pow(2, n);
printf("%d\n", m%x);
return 0;
}