Elegant fibonacci numbers again

Problem Description

Fibonacci numbers are nice simple integers. All of you are familiar with it, aren’t you?

The Fibonacci sequence <F
> are defined by the recurrence:

F[0]=0;

F[1]=1;

F
=F[n-1]+F[n-2], for n>1

You know that the value of F
increases rapidly when the n becomes larger. So for each test case,output the value F
mod m will be ok.

Input

The first line of the input is a positive integer.It is the number of the test cases followed. Each test case contains two integers n (0<=n<2^32) and m (0<m<10000). There may be one or several spaces between the two integers.

Output

The output of the program should consist of one line of output for each test case.The output of each test case only contains one integer which equals to the value F
mod m. No any redundant spaces are needed.

Sample Input

2
1 1000
2 100

Sample Output

1
1

// 关键字:  斐波拉契数列的矩阵求法

//标程：

#include<stdio.h>
struct ss
{
int a[2][2];
};
ss  fun(ss x,ss y,__int64 m)
{
ss  ret;
ret.a[0][0]=(x.a[0][0]*y.a[0][0]%m+x.a[0][1]*y.a[1][0]%m)%m;
ret.a[0][1]=(x.a[0][0]*y.a[0][1]%m+x.a[0][1]*y.a[1][1]%m)%m;
ret.a[1][0]=(x.a[1][0]*y.a[0][0]%m+x.a[1][1]*y.a[1][0]%m)%m;
ret.a[1][1]=(x.a[1][0]*y.a[0][1]%m+x.a[1][1]*y.a[1][1]%m)%m;
return ret;
}
ss f(ss ans,__int64 n,__int64 m)
{
ss ret,temp;
if(n==0)
{
ret.a[0][0]=1,  ret.a[0][1]=0;
ret.a[1][0]=0,  ret.a[1][1]=1;
return ret;
}
if(n==1) return ans;
temp=f(ans,n/2,m);
ret=fun(temp,temp,m);
if(n%2==1) ret=fun(ret,ans,m);
return ret;
}
int main()
{
//freopen("a.txt","r",stdin);
__int64 n,m;
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%I64d%I64d",&n,&m);
if(n==0) { printf("0\n");  continue;}
ss  ans;
ans.a[0][0]=1,  ans.a[0][1]=1;
ans.a[1][0]=1,  ans.a[1][1]=0;
ss  s=f(ans,n-1,m);
printf("%d\n",s.a[0][0]);
}
return 0;
}