/*做题方法： 前是打表，就是打出一大堆数据，然后发现规率：发现这道题是从f[1]=1,和f[2]=1开始，然后依次模7，就可知f[n]只有7种情况，而相数相邻只有7*7=49种，所以从f[1]到f[49]必会出现相邻两个f[m-1]=1,f[m]=1，所以f[m]为周期函数，49为其一个周期;代码如下：*/#include<stdio.h>int main(){    int f[56],a,b,i;    f[0]=1;f[1]=1;    int n;    while(1)    {        scanf("%d%d%d",&a,&b,&n);        if(!a && !b && !n)            break;        for(i=2;i<49;i++)            f[i]=(a*f[i-1]+b*f[i-2])%7;        printf("%d\n",f[(n-1)%49]);    }    return 0;}

Problem Description
A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).
 


 

Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
 


 

Output
For each test case, print the value of f(n) on a single line.
 


 

Sample Input
1 1 3 1 2 10 0 0 0
 


 

Sample Output
2 5