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

看上去是個矩陣快速冪的題目，實際上由於mod 7導致f(n-1)和f(n-2)只有49種可能的情況，必然會迴圈，找規律即可。

#include<iostream>
#include<cstring>
#include<algorithm>
typedef long long ll;
using namespace std;
int f(int A,int B,int n)
{
	if(n==1||n==2)
		return 1;
	else
		return (A*f(A,B,n-1)+B*f(A,B,n-2))%7;

}
int main()
{
	int a,b,n;
	while(cin>>a>>b>>n,a||b||n)
	{
		cout<<f(a,b,n%49)<<endl;
	}
	return 0;
}