URAL 1528Sequence

Description

You are given a recurrent formula for a sequence f:

f( n) = 1 + f(1) g(1) + f(2) g(2) + … + f( n−1) g( n−1),

where g is also a recurrent sequence given by formula

g( n) = 1 + 2 g(1) + 2 g(2) + 2 g(3) + … + 2 g( n−1) − g( n−1) g( n−1).

It is known that f(1) = 1, g(1) = 1. Your task is to find f( n) mod  p.

Input

The input consists of several cases. Each case contains two numbers on a single line. These numbers are n (1 ≤  n ≤ 10000) and p (2 ≤ p ≤ 2·10 ⁹). The input is terminated by the case with n = p = 0 which should not be processed. The number of cases in the input does not exceed 5000.

Output

Output for each case the answer to the task on a separate line.

Sample Input

inputoutput

1 22 110 0

12

题目在吓唬人啊，仔细算一算这个就是阶乘啊。

#include<cstdio>#include<cstring>#include<iostream>#include<string>#include<algorithm>using namespace std;int main(){int n, p;while (cin >> n >> p, n, p){long long tot = 1;for (int i = 1; i <= n; i++) tot = tot*i%p;cout << tot << endl;}return 0;}