Description

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.

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 F_n

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
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:

.

題意：求f[n]，f為斐波那契數列

思路：矩陣快速冪和一般的快速冪是一樣的，只是把普通的乘法改成矩陣乘法即可，可以當模板。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define mod 10000
struct Matrix
{
    long long ma[2][2];
};
Matrix mul(Matrix A,Matrix B)
{
    Matrix C;
    C.ma[0][0]=C.ma[0][1]=C.ma[1][0]=C.ma[1][1]=0;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            for(int k=0;k<2;k++)
            {
                C.ma[i][j]=(C.ma[i][j]+A.ma[i][k]*B.ma[k][j])%mod;
            }
        }
    }
    return C;
}
Matrix pow_mod(Matrix A,long long n)
{
    Matrix B;
    B.ma[0][0]=B.ma[1][1]=1;
    B.ma[0][1]=B.ma[1][0]=0;
    while(n)
    {
        if(n&1) B=mul(B,A);
        A=mul(A,A);
        n>>=1;
    }
    return B;
}
int main()
{
    long long n;
    while(~scanf("%lld",&n)&&n!=-1)
    {
        Matrix A;
        A.ma[0][0]=1;A.ma[0][1]=1;
        A.ma[1][0]=1;A.ma[1][1]=0;
        Matrix ans=pow_mod(A,n);
        printf("%lld\n",ans.ma[0][1]);
    }
    return 0;
}