In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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.

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

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

typedef long long int LL;

using namespace std;

int n,mod=;

struct Matrix

{

    int a[][];

    Matrix(){memset(a,,sizeof(a));}

    Matrix operator* (const Matrix &p)

    {

        Matrix res;

        for(int i=;i<;i++)

        {

            for(int j=;j<;j++)

            {

                for(int k=;k<;k++)

                {

                    res.a[i][j]+=(a[i][k]*p.a[k][j]%mod);

                }

                res.a[i][j]%=mod;

            }

        }

        return res;

    }

}ans,base;

Matrix quick_pow(Matrix base,int k)

{

    Matrix res;

    for(int i=;i<;i++)

    {

        res.a[i][i]=;

    }

    while(k)

    {

        if(k&)  res=res*base;

        base=base*base;

        k>>=;

    }

    return res;

}

void init_Matrix()

{

    ans.a[][]=;

    ans.a[][]=;

    ans.a[][]=;

    ans.a[][]=;

    base.a[][]=;

    base.a[][]=;

    base.a[][]=;

    base.a[][]=;

}

int main()

{

    while(scanf("%d",&n)&&n!=-)

    {

        init_Matrix();

        if(n==) printf("0\n");

        else if(n==) printf("1\n");

        else

        {

            ans=ans*quick_pow(base,n-);

            printf("%d\n",ans.a[][]);

        }

    }

    return ;

}