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).

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 a single integer representing f

2 3
3

1

0 -1
2

1000000006

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).

題意：已知f1,f2,n   f【n】=f【n-1】-f【n-2】， 求f【n】

矩陣快速冪一下！！！

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
struct mat
{
    ll a[2][2];
    mat()
    {
        a[0][0]=a[1][1]=1;
        a[0][1]=a[1][0]=0;
    }
};
mat mul(mat A,mat B)
{
    mat ans;
    int i,j,k;
    memset(ans.a,0,sizeof(ans.a));
    for(i=0;i<2;i++) {
        for(j=0;j<2;j++) {
            for(k=0;k<2;k++) {
                ans.a[i][j]+=(A.a[i][k]*B.a[k][j]%mod+mod)%mod;
                ans.a[i][j]%=mod;
            }
        }
    }
    return ans;
}
mat work(mat A,ll n)
{
    mat ans;
    while(n) {
        if(n&1) ans=mul(A,ans);
        A=mul(A,A);
        n>>=1;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    ll f1,f2,fn,n;
    mat A;
    cin>>f1>>f2>>n;
    f1%=mod,f2%=mod;
    f1+=mod,f2+=mod;
    f1%=mod,f2%=mod;
    A.a[0][0]=A.a[1][0]=1;
    A.a[0][1]=-1,A.a[1][1]=0;
    if(n<3) {
        if(n==1) cout<<f1<<endl;
        if(n==2) cout<<f2<<endl;
        return 0;
    }
    A = work(A,n-2);
    fn = (A.a[0][0]*f2%mod + A.a[0][1]*f1%mod + mod)%mod;
    cout<<fn<<endl;
    return 0;
}