CF--Almost All Divisors--数学基础

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探讨一种算法挑战，通过给定除1和自身外的所有因数，寻找最小可能的整数。解析输入数据，利用排序和乘法验证猜想，最终输出结果或-1表示无法找到该数。

D. Almost All Divisors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=300+66;
const ll mod=1e9+7;
int q,n;
int a[maxn];
bool check(int x,int n)
{
    for(int i=1;i<=n;i++)
    {
        if(x%a[i])return false;
    }
    return true;
}
int main()
{
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        ll x=(ll)a[1]*a[n];
        bool flag=1;
        ll l=1,r=n;
        for(ll i=2;i*i<=x;i++)
        {
            if(x%i==0)
            {
                if(i!=a[l]||x/i!=a[r])flag=0;
                else l++,r--;
            }
        }
        if(l<=r)flag=0;
        printf("%lld\n",flag==0?-1:x);//---//
    }
}