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.
代码：
#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
int quick(int n,int m)
{
int ans=1,base=n;
while(m>0)
{
if(m&1) ans=ans*base%mod;
base=base*base%mod;
m>>=1;
}
return ans;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n>=30)
{
printf("%d\n",m);
continue;
}
printf("%d\n",m%quick(2,n));
}
return 0;
}