D. Almost All Divisors

We guessed some integer number xx. You are given a list of almost all its divisors. Almost all means that there are all divisors except 11and xx in the list.

Your task is to find the minimum possible integer xx that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.

You have to answer tt independent queries.

Input

The first line of the input contains one integer tt (1≤t≤251≤t≤25) — the number of queries. Then tt queries follow.

The first line of the query contains one integer nn (1≤n≤3001≤n≤300) — the number of divisors in the list.

The second line of the query contains nn integers d1,d2,…,dnd1,d2,…,dn (2≤di≤1062≤di≤106), where didi is the ii-th divisor of the guessed number. It is guaranteed that all values didi are distinct.

Output

For each query print the answer to it.

If the input data in the query is contradictory and it is impossible to find such number xx that the given list of divisors is the list of almost allits divisors, print -1. Otherwise print the minimum possible xx.

Example

input

Copy

2
8
8 2 12 6 4 24 16 3
1
2

output

Copy

48
4

题解：对于一个数，他所有的因子是对称分布的，最小的与最大的相乘就为该数，次小的与次大的相乘也为该数，依次类推，那么我们只要根据已知条件求得原来的数是多少，然后再判断该数的所有因子除了1和本身是否都在所给信息中存在并且除了1和本身它的因子个数也应等于给出来的n，判断输出即可。

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int main()
{
    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        vector<ll>v;
        for(int i=;i<=n;i++){
            int x;
            cin>>x;
            v.push_back(x);
        }
        sort(v.begin(),v.end());
        ll ans=v[]*v[v.size()-];
        int sum=,flag=;
        for(ll i=;i*i<=ans;i++){
            if(ans%i==){
                if(find(v.begin(),v.end(),i)==v.end()||find(v.begin(),v.end(),ans/i)==v.end()){flag=;break;}
                sum+=;
                if(i*i==ans)
                    sum--;
            }
        }
        if(sum==n&&flag==)
            cout<<ans<<endl;
        else
            cout<<-<<endl;
    }
    return ;
}