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. 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).

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:

.

题目大意就是求出斐波那契数列第k项的后四位，如果有前导0就不输出0，但最后一个0必须输出

刚开始的时候就想到了一个关于斐波那契数列的数列的一个公式，准备用那个公式求的，但题目已经提示了用矩阵，就没用那个公式

有兴趣的可以看看这个公式，试试能不能那个解这题

这一题就要注意到矩阵

的右上角或左下角对应的值正好是第n项的值

故可以通过快速幂来求得斐波那契数列的哦n项，由于数据较大，还要记得要不断的对10000取模

矩阵的快速幂我是直接抄的模板，刚开始时不理解，现在懂了，详情直接看代码

参考代码;

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 struct prog {
 9     int a[2][2] ;
10     void init(){
11         a[0][0]=a[1][0]=a[0][1]=1;
12         a[1][1]=0;
13     }
14 };
15 
16 prog matrixmul ( prog a ,prog b )
17 {
18     int i , j , k ;
19     prog c ;
20     for ( i = 0 ; i < 2; i ++ )
21     {
22         for ( j = 0 ; j < 2 ; j ++ )
23         {
24             c.a[i][j]=0;
25