Reading comprehension


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 188    Accepted Submission(s): 82


Problem Description


Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>

const int MAX=100000*2;
const int INF=1e9;

int main()
{
  int n,m,ans,i;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    ans=0;
    for(i=1;i<=n;i++)
    {
      if(i&1)ans=(ans*2+1)%m;
      else ans=ans*2%m;
    }
    printf("%d\n",ans);
  }
  return 0;
}


 



Input


Multi test cases,each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000


 



Output


For each case,output an integer,represents the output of above program.


 



Sample Input


1 10 3 100


 



Sample Output


1 5


 


当i是偶数时f[i]=2*f[i-1]  否则f[i]=2*f[i-1]+1
那么我们考虑,f[2*i]=4*f[2*i-2]+2
这样偶项就成为了一个独立的递推数列。
令b[0]=0,b[i]=4*b[i-1]+2
对于f
当i为偶数时,计算b[i/2]就可以了。
当i为奇数时,计算b[i/2]*2+1
计算b[i]可以建立矩阵递推




[4011]


最后右边乘上列向量

(0,2)T
,上面那一项就是b[i]了。