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2k>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 }