#include <bits/stdc++.h>
int comp(const void* a, const void* b) { return *(int*)a - *(int*)b; }
int main() {
  int n, t, ok;
  long long int ans = 0;
  scanf("%d", &t);
  while (t--) {
    int count = 0, ok = 1;
    scanf("%d", &n);
    int a[n];
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    qsort(a, n, 4, comp);
    ans = (long long int)a[0] * a[n - 1];
    for (int i = 1; i < n - 1; i++) {
      if (ans != a[i] * a[n - i - 1]) {
        ok = 0;
        break;
      }
    }
    for (int i = 2; i < sqrt(ans); i++) {
      if (ans % i == 0) count++;
    }
    if ((n % 2 == 0 && count != n / 2) ||
        (n % 2 == 1 && count != (n - 1) / 2) || ok == 0 ||
        (ans == (int)sqrt(ans) * (int)sqrt(ans) && n % 2 == 0))
      ans = -1;
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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>
int comp(const void* a, const void* b) { return *(int*)a - *(int*)b; }
int main() {
  int n, t, ok;
  long long int ans = 0;
  scanf("%d", &t);
  while (t--) {
    int count = 0, ok = 1;
    scanf("%d", &n);
    int a[n];
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    qsort(a, n, 4, comp);
    ans = (long long int)a[0] * a[n - 1];
    for (int i = 1; i < n - 1; i++) {
      if (ans != a[i] * a[n - i - 1]) {
        ok = 0;
        break;
      }
    }
    for (int i = 2; i < sqrt(ans); i++) {
      if (ans % i == 0) count++;
    }
    if ((n % 2 == 0 && count != n / 2) ||
        (n % 2 == 1 && count != (n - 1) / 2) || ok == 0 ||
        (ans == (int)sqrt(ans) * (int)sqrt(ans) && n % 2 == 0))
      ans = -1;
    printf("%I64d\n", ans);
  }
  return 0;
}

```