People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow:

Each player chooses two numbers Ai and Bi and writes them on a
slip of paper. Others cannot see the numbers. In a given moment all
players show their numbers to the others. The goal is to determine the
sum of all expressions Ai
^Bi from all players including oneself and determine
the remainder after division by a given number M. The winner is the one
who first determines the correct result. According to the players'
experience it is possible to increase the difficulty by choosing higher
numbers.

You should write a program that calculates the result and is able to find out who won the game.

The input consists of Z assignments. The number of them is given
by the single positive integer Z appearing on the first line of input.
Then the assignements follow. Each assignement begins with line
containing an integer M (1 <= M <= 45000). The sum will be divided
by this number. Next line contains number of players H (1 <= H <=
45000). Next exactly H lines follow. On each line, there are exactly
two numbers Ai and Bi separated by space. Both numbers cannot be equal
zero at the same time.

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

3

16

4

2 3

3 4

4 5

5 6

36123

1

2374859 3029382

17

1

3 18132

2

13195

13


题目大意：给一个M， 再给一个n表示接下来有n组数据(a,b) 计算a的b次幂，在将这n组数据加在一起。 然后对M求余。
快速幂求余，a^b%m=[(a%m)^b]%m
同余定理  (a+b+c...)%m=(a%m+b%m+c%m...)%m
AC代码：

#include<iostream>

#include<cstdio>

using namespace std;

typedef long long ll;

ll m;//模

int pow(ll x,ll y)

{

    ll res=;

    while(y)

    {

        if(y&)

            res=res*x%m;

        x=x*x%m;

        y>>=;

    }

    return res%m;//(a+b+c...)%m=(a%m+b%m+c%m..)%m

 } 

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        ll c;

        scanf("%lld %lld",&m,&c);

        ll a,b,sum=;

        for(int i=;i<c;i++)

        {

            scanf("%lld %lld",&a,&b);

            sum+=pow(a,b);

        }

        printf("%lld\n",sum%m);

     }

    return ;

 }