CodeForces 450B

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).

 

/**

          题意：f[i] = f[i-1] + f[i+1]

          做法：矩阵  如题要求建一个二维矩阵，

                    0   1

                   -1   0

                   然后求解

**/

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

#define SIZE 2

#define MOD 1000000007

#define clr( a, b ) memset( a, b, sizeof(a) )

typedef long long LL;



struct Mat

{

    LL mat[ SIZE ][ SIZE ];

    int n;

    Mat( int _n )

    {

        n = _n;

        clr( mat, 0 );

    }

    void init()

    {

        for( int i = 0; i < n; ++i )

            for( int j = 0; j < n; ++j )

                mat[i][j] = ( i == j );

    }

    Mat operator * ( const Mat &b ) const

    {

        Mat c( b.n );

        for( int k = 0; k < n; ++k )

            for( int i = 0; i < n; ++i )    if( mat[i][k] )

                    for( int j = 0; j < n; ++j )

                        c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * b.mat[k][j] ) % MOD;

        return c;

    }

};



Mat fast_mod( Mat a, int b )

{

    Mat res( a.n );

    res.init();

    while( b )

    {

        if( b & 1 ) res = res * a;

        a = a * a;

        b >>= 1;

    }

    return res;

}





int main()

{

    LL x, y, n, res;

    scanf( "%lld %lld %lld", &x, &y, &n );

    if( n == 1 )

    {

        printf( "%lld\n", ( x % MOD + MOD ) % MOD );

    }

    else if( n == 2 )

    {

        printf( "%lld\n", ( y % MOD + MOD ) % MOD );

    }

    else

    {

        n -= 2;

        Mat C( 2 );

        C.mat[0][0] = 0;

        C.mat[0][1] = 1;

        C.mat[1][0] = -1;

        C.mat[1][1] = 1;

        C = fast_mod( C, n );

        res = ( ( x * C.mat[1][0] + y * C.mat[1][1] )% MOD + MOD ) % MOD;

        printf( "%lld\n", res );

    }

    return 0;

}