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

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:

.

PS：基本的矩陣快速冪

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
const int maxn=1000000;
const int mod=10000;
#define me(a) memset(a,0,sizeof(a))
#define ll long long
using namespace std;
struct node{
    int a[2][2];
    node()
    {
        memset(a,0,sizeof(a));
    }
};
node jx(node a,node b)
{
    node s;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                s.a[i][k]=(s.a[i][k]+a.a[i][j]*b.a[j][k])%mod;
    return s;
}
node qsort(node a,int n)
{
    node s;
    for(int i=0;i<2;i++)
        s.a[i][i]=1;
    while(n)
    {
        if(n&1)
            s=jx(s,a);
        a=jx(a,a);
        n>>=1;
    }
    return s;
}
int main()
{
    node a;
    int n;
    a.a[0][0]=a.a[1][0]=a.a[0][1]=1;
    while(cin>>n&&n!=-1)
    {
        node s=qsort(a,n);
        cout<<s.a[1][0]<<endl;
    }
    return 0;
}

。