Jzzhu and Sequences CodeForces - 450B 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 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

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).

思路：构建矩阵

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll x,y;
ll f[2][2];
ll tmp[2][2];
ll res[2][2],b[2][2];
void multi(ll f[][2],ll b[][2],ll n)
{memset(tmp,0,sizeof(tmp));for(int i=0;i<2;i++)for(int j=0;j<2;j++)for(int k=0;k<2;k++)tmp[i][j]=((tmp[i][j]+f[i][k]*b[k][j])%mod+mod)%mod;for(int i=0;i<2;i++)for(int j=0;j<2;j++)f[i][j]=tmp[i][j];
}
void Pow(ll n)
{memset(res,0,sizeof(res));for(int i=0;i<2;i++)res[i][i]=1;while(n){if(n&1)multi(res,f,n);multi(f,f,n);n>>=1;}
}
int main()
{ll n;while(~scanf("%lld%lld",&x,&y)){f[0][0]=f[1][0]=1;f[0][1]=-1;f[1][1]=0;scanf("%lld",&n);if(n==1){if(x<0){printf("%lld\n",x+mod);}elseprintf("%lld\n",x);continue;}Pow(n-2);ll ans;ans=res[0][0]*y%mod+res[0][1]*x%mod;printf("%lld\n",(ans%mod+mod)%mod);}return 0;
}

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