#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    long long mn = 1e9, mx = -1e9;
    long long a[n];
    for (int j = 0; j < n; j++) {
      cin >> a[j];
      mn = min(a[j], mn);
      mx = max(a[j], mx);
    }
    long long x = mn * mx;
    int ans = 1;
    set<long long> s;
    for (int j = 2; j <= sqrt(x); j++) {
      if (x % j == 0) {
        s.insert(j);
        s.insert(x / j);
      }
    }
    int cnt = 0;
    for (int j = 0; j < n; j++) {
      if (x % a[j] == 0) {
        cnt++;
      } else {
        cnt = -1;
        break;
      }
    }
    if (cnt != s.size()) {
      cout << -1 << endl;
    } else
      cout << x << endl;
  }
}

### 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    long long mn = 1e9, mx = -1e9;
    long long a[n];
    for (int j = 0; j < n; j++) {
      cin >> a[j];
      mn = min(a[j], mn);
      mx = max(a[j], mx);
    }
    long long x = mn * mx;
    int ans = 1;
    set<long long> s;
    for (int j = 2; j <= sqrt(x); j++) {
      if (x % j == 0) {
        s.insert(j);
        s.insert(x / j);
      }
    }
    int cnt = 0;
    for (int j = 0; j < n; j++) {
      if (x % a[j] == 0) {
        cnt++;
      } else {
        cnt = -1;
        break;
      }
    }
    if (cnt != s.size()) {
      cout << -1 << endl;
    } else
      cout << x << endl;
  }
}

```