#include <bits/stdc++.h>
using namespace std;
const int N = 10e6 + 100;
int main() {
  int t;
  long long n;
  scanf("%d", &t);
  while (t--) {
    int k = 0;
    scanf("%lld", &n);
    set<long long> Start, End;
    vector<long long> st, en;
    long long maxx = 1;
    long long minn = N;
    for (long long i = 0; i < n; i++) {
      long long a;
      scanf("%lld", &a);
      Start.insert(a);
      maxx = max(maxx, a);
      minn = min(minn, a);
    }
    long long newNumber = maxx * minn;
    for (long long i = 2; i * i <= newNumber; i++) {
      if (newNumber % i == 0) {
        End.insert(i);
        if (i != newNumber / i) {
          End.insert(newNumber / i);
        }
      }
    }
    for (set<long long>::iterator f = Start.begin(); f != Start.end(); f++) {
      st.push_back(*f);
    }
    for (set<long long>::iterator f = End.begin(); f != End.end(); f++) {
      en.push_back(*f);
    }
    if (st == en) {
      cout << newNumber << "\n";
    } else {
      cout << "-1"
           << "\n";
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

We guessed some integer number x. You are given a list of almost all its divisors. Almost all means that there are all divisors except 1 and 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 (1 ≤ t ≤ 25) — the number of queries. Then t queries follow.

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

The second line of the query contains n integers d_1, d_2, ..., d_n (2 ≤ d_i ≤ 10^6), where d_i is the i-th divisor of the guessed number. It is guaranteed that all values d_i 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 x that the given list of divisors is the list of almost all its divisors, print -1. Otherwise print the minimum possible x.

Example

Input


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


Output


48
4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 10e6 + 100;
int main() {
  int t;
  long long n;
  scanf("%d", &t);
  while (t--) {
    int k = 0;
    scanf("%lld", &n);
    set<long long> Start, End;
    vector<long long> st, en;
    long long maxx = 1;
    long long minn = N;
    for (long long i = 0; i < n; i++) {
      long long a;
      scanf("%lld", &a);
      Start.insert(a);
      maxx = max(maxx, a);
      minn = min(minn, a);
    }
    long long newNumber = maxx * minn;
    for (long long i = 2; i * i <= newNumber; i++) {
      if (newNumber % i == 0) {
        End.insert(i);
        if (i != newNumber / i) {
          End.insert(newNumber / i);
        }
      }
    }
    for (set<long long>::iterator f = Start.begin(); f != Start.end(); f++) {
      st.push_back(*f);
    }
    for (set<long long>::iterator f = End.begin(); f != End.end(); f++) {
      en.push_back(*f);
    }
    if (st == en) {
      cout << newNumber << "\n";
    } else {
      cout << "-1"
           << "\n";
    }
  }
  return 0;
}

```