codeforces 913A Modular Exponentiation

codeforces 913A Modular Exponentiation

​​ 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.

模拟。怎么搞 看一下我给的2^n是否大于m大于就输出m否则 算出2^n 然后取模即可

#include #includelong long n,m;int main(){ freopen("cf.in","r",stdin); scanf("%lld",&n);scanf("%lld",&m);long long base=1;int now=0; for (int i=1;i<=n;++i){ base*=2;if(base>m) {now=i;break;} } if (!now) printf("%lld",m%base);else{ printf("%lld",m); } return 0;}

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