#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res;
  if (b == 0) return 1;
  if (b == 1) return a % 1000000007;
  if (b & 1) {
    res = power(a, b - 1) % 1000000007;
    res = ((a % 1000000007) * (res % 1000000007)) % 1000000007;
  } else {
    res = power(a, b / 2) % 1000000007;
    res = ((res % 1000000007) * (res % 1000000007)) % 1000000007;
  }
  return res;
}
void solve() {
  long long n, i, a[200005], num, mn = 1e10, mx = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(mn, a[i]);
    mx = max(mx, a[i]);
  }
  num = mn * mx;
  vector<long long> v;
  for (i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      v.push_back(i);
      if (i != sqrt(num)) v.push_back(num / i);
    }
  }
  if (n != v.size())
    cout << "-1"
         << "\n";
  else {
    sort(a, a + n);
    sort(v.begin(), v.end());
    for (i = 0; i < n; i++) {
      if (a[i] != v[i]) {
        cout << "-1"
             << "\n";
        return;
      }
    }
    cout << num << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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>
using namespace std;
long long power(long long a, long long b) {
  long long res;
  if (b == 0) return 1;
  if (b == 1) return a % 1000000007;
  if (b & 1) {
    res = power(a, b - 1) % 1000000007;
    res = ((a % 1000000007) * (res % 1000000007)) % 1000000007;
  } else {
    res = power(a, b / 2) % 1000000007;
    res = ((res % 1000000007) * (res % 1000000007)) % 1000000007;
  }
  return res;
}
void solve() {
  long long n, i, a[200005], num, mn = 1e10, mx = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(mn, a[i]);
    mx = max(mx, a[i]);
  }
  num = mn * mx;
  vector<long long> v;
  for (i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      v.push_back(i);
      if (i != sqrt(num)) v.push_back(num / i);
    }
  }
  if (n != v.size())
    cout << "-1"
         << "\n";
  else {
    sort(a, a + n);
    sort(v.begin(), v.end());
    for (i = 0; i < n; i++) {
      if (a[i] != v[i]) {
        cout << "-1"
             << "\n";
        return;
      }
    }
    cout << num << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```