CodeForces - 450B Jzzhu and Sequences (矩阵快速幂)

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

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

Examples

Input

2 3
3

Output

1

Input

0 -1
2

Output

1000000006

由题目中给的 F(i)=F(i-1)+F(i+1) 推导可得 F(i)=F(i-1)-F(i-2)

跟斐波那契数列类似，矩阵快速幂，构造矩阵为  {1,-1,1,0}

#include<bits/stdc++.h>
using namespace std;
const int maxn=3000+5;
const int mod=1e9+7;
typedef long long ll;
ll x,y,n;
struct Matrix {
  ll a[2][2];
  Matrix() { memset(a, 0, sizeof (a)); }
  Matrix operator*(const Matrix &b) const {
    Matrix res;
    for (int i = 0; i <2; ++i)
      for (int j = 0; j <2; ++j)
        for (int k = 0; k < 2; ++k)
          res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j]) % mod;
    return res;
  }
} ans, base;

void init() {
  base.a[0][0]  = base.a[1][0] = 1;
  base.a[0][1]=-1; 
  ans.a[0][0] = ans.a[1][1] = 1;
}

void qpow(int b) {
  while (b) {
    if (b & 1) ans = ans * base;
    base = base * base;
    b >>= 1;
  }
}

int main(int argc, char const *argv[])
{
	scanf("%lld%lld",&x,&y);
	scanf("%lld",&n);
	if(n==1) printf("%lld\n",(x%mod+mod)%mod);
	else if(n==2) printf("%lld\n",(y%mod+mod)%mod);
	else
	{
		init();
		qpow(n-2);
		ll res=(((ans.a[0][0]*y+ans.a[0][1]*x)%mod)+mod)%mod;
		printf("%lld\n",res);
	} 
	return 0;
}