Almost All Divisors（求因子个数及思维）

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We guessed some integer number x in the list.

Your task is to find the minimum possible integer x 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 t independent queries.

Input

The first line of the input contains one integer t queries follow.

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

The second line of the query contains 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 x.

Example

input

Copy

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

output

Copy

48
4

 

 

 

思路：求出因子个数，看是否这n个数是否包含这n个因子数，然后判断一下再判断一下这n个数是否是他的因子

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
ll count(ll n){
    ll s=1;
    for(ll i=2;i*i<=n;i++){
        if(n%i==0){
            int a=0;
            while(n%i==0){
                n/=i;
                a++;
            }
            s=s*(a+1);
        }
    }
    if(n>1) s=s*2;
    return s;
}
ll a[maxn];
int main()
{
   int T;
   cin>>T;
   int n;
   while(T--)
   {
       scanf("%d",&n);
       ll maxx=2;
       ll minn=10000000;
       ll x;
       for(int t=0;t<n;t++)
       {
           scanf("%lld",&a[t]);
           maxx=max(a[t],maxx);
           minn=min(a[t],minn);
    }
    ll ans=maxx*minn;
    bool flag=false;
    for(int t=0;t<n;t++)
    {
        if(ans%a[t]!=0)
        {
            flag=true;
        }
    }
    if(count(ans)-2==n&&flag==false)
    printf("%lld\n",ans);
    else
    {
        printf("-1\n");
    }
    

   }
   return 0;
}

 

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