Modular Exponentiation (思维)

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本文介绍了一个计算m除以2的n次方取余数的问题，并提供了一种简单高效的解决方法。对于n大于30的情况，直接返回m作为结果；而对于n小于等于30的情况，则通过直接计算得到结果。

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

$\text{[math]}$ ,

where 2ⁿ = 2·2·...·2 (n factors), and $\text{[math]}$ denotes the remainder of division of x by y.

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

$\text{[math]}$ .

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 $\text{[math]}$ .

Examples

Input

4
42

Output

Input

1
58

Output

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：求m%(2^n)
思路：我们知道2^30大于10^8,所以当n>30时结果一定是m
而当n<30时，2^n 不会超long long,所以直接求即可
代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    ll n,m;
    cin>>n>>m;
    if(n>=30) cout<<m<<endl;
    else cout<<m%(1<<n)<<endl;
	return 0;
}