F - Modular Exponentiation

Problem description

The following problem is well-known: given integers n and m, calculate 2ⁿ mod m,

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

You are asked to solve the "reverse" problem. Given integers n and m, calculate m mod 2ⁿ.

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 m mod 2ⁿ.

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.

解题思路：由于给出的m最大值为10⁸，于是暴力找出2^k>10⁸时的最小值k，解得k=27，所以只要n>26，直接输出m（取模一个比自己大的数字，结果为本身），反之直接取模运算，这样就不会发生数据溢出。（位运算是个好东西，长记性了）

AC代码：

1 #include <bits/stdc++.h>
2 using namespace std;
3 int main()
4 {
5     int n,m;
6     cin>>n>>m;
7     cout<<(n>26?m:m%(1<<n))<<endl;
8     return 0;
9 }

 

转载于:https://www.cnblogs.com/acgoto/p/9103203.html