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

.<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+R2l2ZW4gYW4gaW50ZWdlciA8ZW0+bjwvZW0+LCB5b3VyIGdvYWwgaXMgdG8gY29tcHV0ZSB0aGUgbGFzdCA0IGRpZ2l0cyBvZiA8ZW0+RjxzdWI+bjwvc3ViPjwvZW0+LjwvcD4KCgoKCjxwIGNsYXNzPQ=="pst"> 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:

.

题意：题意：求第n个Fibonacci数mod(m)的结果，当n=-1时，break。其中n(where 0 ≤ n ≤ 1,000,000,000) ，m=10000；

思路：常规方法肯定超时，这道题学会了用矩阵快速幂求斐波那契。如下图：

A = F(n - 1), B = F(N - 2)，这样使构造矩阵的n次幂乘以初始矩阵得到的结果就是。

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