杭电Questionnaire

Problem Description

In order to get better results in official ACM/ICPC contests, the team leader comes up with a questionnaire. He asked everyone in the team whether to have more training.



Picture from Wikimedia Commons


Obviously many people don't want more training, so the clever leader didn't write down their words such as ''Yes'' or ''No''. Instead, he let everyone choose a positive integerai to represent his opinion. When finished, the leader will choose a pair of positive intergesm(m>1) and k(0≤k<m), and regard those people whose number is exactly k modulo m as ''Yes'', while others as ''No''. If the number of ''Yes'' is not less than ''No'', the leader can have chance to offer more training.

Please help the team leader to find such pair of m and k.

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there is an integer n(3≤n≤100000) in the first line, denoting the number of people in the ACM/ICPC team.

In the next line, there are n distinct integers a1,a2,...,an(1≤ai≤109), denoting the number that each person chosen.

 

Output

For each test case, print a single line containing two integersm and k, if there are multiple solutions, print any of them.

 

Sample Input

1623 3 18 8 13 9

 

Sample Output

5 3
题意：给你n个数，让你找一对数，m,k，使得这n个数 模上m等于k的个数 大于等于 模上m不等于k的个数   m>1  0<=k<m
思路：所有的数模上2之后为0或者1
AC代码如下：
#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int maxn=1e5+10;int a[maxn];int main(){    int t;    int n;    cin>>t;    while(t--)    {        cin>>n;        int num=0;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            if(a[i]%2==1) num++;        }        if(num>=n-num) cout<<2<<" "<<1<<endl;        else            cout<<2<<" "<<0<<endl;    }    return 0;}