#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
using namespace std;
const double EPS = 0.0000000001;
long long int get(long long int n) {
  long long int c = 1;
  long long int k = sqrt(n);
  for (int i = 2; i <= k && n > 1; i++) {
    if (n % i == 0) {
      long long int cnt = 0;
      while (n > 1 && n % i == 0) {
        cnt++;
        n /= i;
      }
      c *= (cnt + 1);
    }
  }
  if (n > 1) c *= 2;
  return c - 2;
}
bool func(long long int n) {
  if (n == 1) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (int i = 3; i <= sqrt(n + 1); i += 2) {
    if (!(n % i)) return 0;
  }
  return 1;
}
int main() {
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    long long int n;
    cin >> n;
    long long int mass[n];
    for (int z = 0; z < n; z++) cin >> mass[z];
    sort(mass, mass + n);
    long long int c = mass[0] * mass[n - 1];
    bool p = 0;
    if (!func(mass[0])) {
      cout << -1 << endl;
      continue;
    }
    for (int z = 0; z < n; z++) {
      if (c % mass[z]) {
        p = 1;
        break;
      }
      if (!binary_search(mass, mass + n, c / mass[z])) {
        p = 1;
        break;
      }
    }
    if (p) {
      cout << -1 << endl;
      continue;
    }
    if (get(c) != n) {
      cout << -1 << endl;
      continue;
    }
    cout << c << endl;
  }
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
using namespace std;
const double EPS = 0.0000000001;
long long int get(long long int n) {
  long long int c = 1;
  long long int k = sqrt(n);
  for (int i = 2; i <= k && n > 1; i++) {
    if (n % i == 0) {
      long long int cnt = 0;
      while (n > 1 && n % i == 0) {
        cnt++;
        n /= i;
      }
      c *= (cnt + 1);
    }
  }
  if (n > 1) c *= 2;
  return c - 2;
}
bool func(long long int n) {
  if (n == 1) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (int i = 3; i <= sqrt(n + 1); i += 2) {
    if (!(n % i)) return 0;
  }
  return 1;
}
int main() {
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    long long int n;
    cin >> n;
    long long int mass[n];
    for (int z = 0; z < n; z++) cin >> mass[z];
    sort(mass, mass + n);
    long long int c = mass[0] * mass[n - 1];
    bool p = 0;
    if (!func(mass[0])) {
      cout << -1 << endl;
      continue;
    }
    for (int z = 0; z < n; z++) {
      if (c % mass[z]) {
        p = 1;
        break;
      }
      if (!binary_search(mass, mass + n, c / mass[z])) {
        p = 1;
        break;
      }
    }
    if (p) {
      cout << -1 << endl;
      continue;
    }
    if (get(c) != n) {
      cout << -1 << endl;
      continue;
    }
    cout << c << endl;
  }
}

```