A

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 are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

/*直接套快速幂模板，求矩阵的n次幂int ans=1;while(n){    if(n&1)        ans*=x;    x*=x;    n>>=1;就是此模板啦，将相应的ans 和x换成矩阵就行*/#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int n,a[2][2],ans[2][2];void mul(int a[2][2],int b[2][2]){    int temp[2][2];    memset(temp,0,sizeof temp);    for(int i=0;i<2;i++){        for(int j=0;j<2;j++){            for(int k=0;k<2;k++){                temp[i][j]=(temp[i][j]+a[i][k]*b[k][j])%10000;            }        }    }    for(int i=0;i<2;i++){        for(int j=0;j<2;j++){            b[i][j]=temp[i][j];        }    }}int main(){    while(scanf("%d",&n)!=EOF){        if(n==-1)           return 0;        a[0][0]=a[0][1]=a[1][0]=1;        a[1][1]=0;        ans[0][0]=ans[1][1]=1;        ans[0][1]=ans[1][0]=0;//相当于ans=1,初始化为单位矩阵        while(n){            if(n&1)                mul(a,ans);//此过程相当于求ans=ans*a;            mul(a,a);//相当于求a=a*a;            n>>=1;        }        printf("%d\n",ans[0][1]);    }}