学步园

B. Jzzhu and Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Sample test(s)

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).

/*
一般碰到这样题基本上就是找规律了
可以开个10000个数组打个表看下规律
此题规律就是每6个循环：前面3个和后面3个正好是相反数
另外此题要注意当n正好是6的倍数要特判下还有一定先取模再想减不然会爆掉
*/

#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
const int Max=10000;
long long f[Max];
long long x[Max];
using namespace std;
/*void init(long long x,long long y)
{
    long long i;
    f[1]=x;
    f[2]=y;
    for(i=2;i<Max;i++)
    {
        f[i+1]=(f[i]-f[i-1])%1000000007;
    }
    for(i=1;i<=1000;i++)
    {
        if(i%20==0)
            cout<<endl;
        cout<<f[i]<<" ";
    }
    cout<<endl;
}*/
int main()
{
    long long x,y;
    long long i,j,n;
    //while(1)
    //{
        cin>>x>>y;
        //init(x,y);
        f[1]=x%1000000007;
        f[2]=y%1000000007;
        f[3]=(f[2]-f[1])%1000000007;
        //cout<<f[3]<<endl;
        f[4]=-f[1];
        f[5]=-f[2];
        f[6]=-f[3];
        cin>>n;
        n%=6;
        if(n==0)
        {
            //cout<<f[6]<<"asd"<<endl;
            if(f[6]<0)
                cout<<(f[6]+1000000007)<<endl;
            else
                cout<<f[6]<<endl;
        }
        else
        {
            if(f[n]<0)
            cout<<(f[n]+1000000007)<<endl;
            else
            cout<<f[n]<<endl;
        }
    //}
    return 0;
}