快速幂

int add(int a, int b){return a + b < mod ? a + b : a + b - mod;}
int mpow(int a, int b)
{
    int c = 1;
    for(;b;b>>=1, a = mul(a, a)){
        if(b & 1) c = mul(a ,c);
    }
    return c;
}