Codeforces450 B. Jzzhu and Sequences

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).

#include <iostream>
#include <cstring>
using namespace std;
/*矩阵快速幂*/
#define mod 1000000007
#define ll long long int
#define sec 2
struct mat
{
    ll arr[sec][sec];
};
mat mul(mat a, mat b)
{
    mat res;
    memset(res.arr, 0, sizeof(res.arr));
    for (int i = 0; i < sec; i++)
    for (int j = 0; j < sec; j++)
    for (int k = 0; k < sec; k++)
    {
        res.arr[i][j] = ((res.arr[i][j] + a.arr[i][k] * b.arr[k][j]) % mod + mod) % mod;
    }
    return res;
}
mat expo(mat ori, int n)
{
    mat res;
    memset(res.arr, 0, sizeof(res.arr));
    for (int i = 0; i < sec; i++)
        res.arr[i][i] = 1;
    while (n)
    {
        if (n & 1)
            res = mul(res, ori);
        ori = mul(ori, ori);
        n >>= 1;
    }
    return res;
}
int main()
{
    int x, y, n;
    while (cin >> x >> y >> n)
    {
        x = (x + mod) % mod;
        y = (y + mod) % mod;
        if (n == 1)
            cout << x << endl;
        else
        if (n == 2)
            cout << y << endl;
        else
        {
            mat a;
            a.arr[0][0] = a.arr[1][0] = 1;
            a.arr[0][1] = -1;
            a.arr[1][1] = 0;
            a = expo(a, n - 2);
            cout << ((a.arr[0][0] * y) % mod + (a.arr[0][1] * x) % mod) % mod << endl;
        }
    }
    return 0;
}