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

input

output

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;
}