As we all know, the next Olympic Games will be held in Beijing in 2008. So the year 2008 seems a little special somehow. You are looking forward to it, too, aren't you? Unfortunately there still are months to go. Take it easy. Luckily you meet me. I have a problem for you to solve. Enjoy your time. 

Now given a positive integer N, get the sum S of all positive integer divisors of 2008 N. Oh no, the result may be much larger than you can think. But it is OK to determine the rest of the division of S by K. The result is kept as M. 

Pay attention! M is not the answer we want. If you can get 2008 M, that will be wonderful. If it is larger than K, leave it modulo K to the output. See the example for N = 1，K = 10000: The positive integer divisors of 20081 are 1、2、4、8、251、502、1004、2008，S = 3780, M = 3780, 2008 M % K = 5776. 
 

Input

The input consists of several test cases. Each test case contains a line with two integers N and K (1 ≤ N ≤ 10000000, 500 ≤ K ≤ 10000). N = K = 0 ends the input file and should not be processed. 

Output

For each test case, in a separate line, please output the result. 

Sample Input

1  10000
0  0

Sample Output

5776

题意：求2008^n的所有因子和s对k取余的值作为m，然后求2008^m对k取余。

题解：注意：不能求逆元，因为不能保证gcd(250,k)=1.请看另一种方法：公式为：x/y%k=x%(k*y)/y 做法：用求因子和的公式，2008=2^3*251，则：2008^n=2^3n*251^n所以因子和为：s=(2^(3^n+1)-1)*(251^(n+1)-1)/250

那么m=s%k=(2^(3^n+1)-1)*(251^(n+1)-1)%(250*k)/250,请看代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
long long quick(long long a,long long b,long long c){//快速幂模板
	long long ans=1;
	a=a%c;
	while(b!=0){
		if(b&1) ans=(ans*a)%c;
		b>>=1;
		a=(a*a)%c;
	}
	return ans;
}
int main(){
	long long n,k,t,m;
	while(cin >> n >> k,n+k){
		k=k*250;//推的公式
		t=(quick(2,3*n+1,k)-1)*(quick(251,n+1,k)-1);//推的公式
		m=t%k/250;//推的公式
		k=k/250;//注意k是原始值，所以要除以250回到原始值
		cout << quick(2008,m,k) << endl;
	}
	return 0;
}