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

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 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

Copy

2 3
3

output

Copy

1

input

Copy

0 -1
2

output

Copy

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：定义一个序列fn,f[1]=x,f[2]=y,f[n]=f[n-1]-f[n-2](n>2)，求这个序列的第n(n<=1e9)项mod1e9+7。

思路：明显是类似于斐波那契数列的矩阵快速幂。

矩阵为，前两项特判即可，注意取模。

代码：

#include
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1002;
ll mo=1000000007;
ll n,a,b;
struct node
{
    ll n,m,a[7][7];
    void init()
    {
        memset(a,0,sizeof(a));
        for(int i=0;i>=1;
        a=mul(a,a,a.n,a.n,a.n);
    }
    return d;
}
ll poww(ll a,ll n)
{
    ll sum=1;
    while(n)
    {
        if(n&1) sum=sum*a%mo;
        n>>=1;
        a=a*a%mo;
    }
    return sum;
}
int main()
{
    int T,cas=1,fg;
    while(scanf("%lld%lld%lld",&a,&b,&n)!=EOF)
    {
        a=(a+mo)%mo;
        b=(b+mo)%mo;
        if(n==1)
        {
            printf("%lld\n",a%mo);
            continue;
        }
        else if(n==2)
        {
            printf("%lld\n",b%mo);
            continue;
        }

        node x;
        x.n=2;x.m=2;
        memset(x.a,0,sizeof(x.a));
        x.a[0][0]=1;
        x.a[0][1]=-1;
        x.a[1][0]=1;
        x.a[1][1]=0;
        node ans=power(x,n-2);

        node tmp;
        tmp.n=2;
        tmp.m=1;
        memset(tmp.a,0,sizeof(tmp.a));
        tmp.a[0][0]=b;
        tmp.a[1][0]=a;

        ans=mul(ans,tmp,2,2,1);
        ll sum=(ans.a[0][0]+mo)%mo;

        printf("%lld\n",sum);
    }
    return 0;
}