Description


In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of Fn.


Input


The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.


Output


For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).


Sample Input


09 999999999 1000000000 -1


Sample Output


034 626 6875


Hint


As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.


第一次学矩阵快速幂，主要用在递推公式上，能大大的提高运行效率，将MOD增大能做大数。

矩阵的幂题目里有说，代码是实现。

矩阵快速幂复杂度一般是 O（lgn）

跑了0ms

代码如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define M 10000
struct edg{
    int  m[2][2];
};
edg matlab(edg x,edg y)
{
    edg ans;
        for(int i=0;i<2;i++)
           for(int j=0;j<2;j++)
         {
        ans.m[i][j]=0;
            for(int k=0;k<2;k++)
                    ans.m[i][j]=(M+x.m[i][k]*y.m[k][j]+ans.m[i][j])%M;
        }
       return ans;
}
int  fast_matlab(int  n)
{
    edg ans,base;
    base.m[0][0]=base.m[0][1]=base.m[1][0]=1;
    base.m[1][1]=0;
    ans.m[0][0]=ans.m[1][1]=1;
    ans.m[1][0]=ans.m[0][1]=0;
    while(n)
    {
        if(n&1)
        ans=matlab(ans,base);
        
        base=matlab(base,base);
        n>>=1;        
    }
    return ans.m[0][1];
}
int main()
{
        int  a,t,n;
        while(cin>>n)
        {
            if(n==-1) break;
            cout<<fast_matlab(n)<<endl;
          }
      return 0;
}