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Problem Description

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 2008N. 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 2008M, 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, 2008M % 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：

先求出2008的n次方的因子和m， 然后再求2008的m次幂取余k。 。。

代码如下：

/*2008=2^3*251;由于250和k不一定互素，所以不能求出250与k的逆元...由于x/d%m=x%(d*m)/d;所以可以对(250*k)进行取余...最后不要忘了/250;求出因子和之后就可以用快速幂取模求出2008的m次方...*/#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;int n,k;long long int fastpow(long long int a,long long int b,long long int c){    long long int sum=1;    while (b>0)    {        if(b%2)            sum=sum*a%c;        a=a*a%c;        b>>=1;    }    return sum;}int main(){    while (scanf("%d%d",&n,&k)!=EOF&&(n||k))    {        long long int a1=fastpow(2,3*n+1,250*k)-1;        long long int b1=fastpow(251,n+1,250*k)-1;        long long int m=a1*b1%(250*k)/250;        printf("%lld\n",fastpow(2008,m,k));    }    return 0;}
      
     
    
   

 

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