Codeforces450 B. Jzzhu and Sequences (找规律)

题目链接：https://vjudge.net/problem/CodeForces-450B

Jzzhu has invented a kind of sequences, they meet the following property:

You are given x and y, please calculate f_n modulo 1000000007 (10⁹ + 7).

Input

The first line contains two integers x and y (|x|, |y| ≤ 10⁹). The second line contains a single integer n (1 ≤ n ≤ 2·10⁹).

Output

Output a single integer representing f_n modulo 1000000007 (10⁹ + 7).

Example

Input

2 3
3

Output

1

Input

0 -1
2

Output

1000000006

Note

In the first sample, f₂ = f₁ + f₃, 3 = 2 + f₃, f₃ = 1.

In the second sample, f₂ =  - 1;  - 1 modulo (10⁹ + 7) equals (10⁹ + 6).

解题思路：

　　由数学公式f₂ = f₁ + f_{3输出任意 fn，看似简单，可是由于数据范围为2e9，所以明显是道矩阵快速幂，可是写的时候推导了前8项，发现每6个一个循环节，于是有了一种更简单的写法，具体看代码吧！}

AC代码：

#include <stdio.h>
#include <math.h>
const long long MOD=1e9+;
int main()
{
    int a[],n;
    while(~scanf("%d %d",&a[],&a[]))
    {
        scanf("%d",&n);
        for(int i=;i<=;i++)
            a[i]=a[i-]-a[i-];
        a[]=a[];
        int ans;
        ans=( a[n%]%MOD + MOD )  %MOD;
        printf("%d\n",ans);

    }
    return ;
}