#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int r = power(a, b / 2) % 1000000007;
  if (b % 2 == 0)
    return (r * r) % 1000000007;
  else
    return (((r * r) % 1000000007) * a) % 1000000007;
}
int prime[1000009];
void sieve(int n) {
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (!prime[i]) {
      for (int j = 2 * i; j <= n; j++) {
        prime[j]++;
      }
    }
  }
}
bool comp(const pair<long long int, long long int> &a,
          const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
const int N = 5 * 1e5 + 9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    map<long long int, long long int> mp1;
    mp1.clear();
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      mp1[a[i]]++;
    }
    if (n == 1) {
      long long int val = a[0];
      int flag = 0;
      for (int i = 2; i <= sqrt(val); i++) {
        if (val % i == 0) {
          flag++;
          continue;
        }
      }
      if (!flag) {
        long long int r = val * val;
        cout << r << "\n";
        continue;
      } else {
        cout << "-1\n";
        continue;
      }
    }
    sort(a, a + n);
    int val = n / 2;
    map<long long int, long long int> mp;
    mp.clear();
    for (int i = 0; i < val; i++) {
      long long int w = a[i] * a[n - i - 1];
      mp[w]++;
    }
    if (mp.size() != 1) {
      cout << "-1\n";
      continue;
    }
    long long int ww = mp.begin()->first;
    if (n % 2) {
      long long int rr = a[val] * a[val];
      if (rr != ww) {
        cout << "-1\n";
        continue;
      }
    }
    int flag = 0;
    for (int i = 2; i <= sqrt(ww); i++) {
      if (ww % i == 0) {
        if (mp1.find(i) == mp1.end()) {
          flag++;
          break;
        }
        if (i * i != ww) {
          long long int kk = ww / i;
          if (mp1.find(kk) == mp1.end()) {
            flag++;
            break;
          }
        }
      }
    }
    if (flag) {
      cout << "-1\n";
    } else
      cout << ww << "\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;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int r = power(a, b / 2) % 1000000007;
  if (b % 2 == 0)
    return (r * r) % 1000000007;
  else
    return (((r * r) % 1000000007) * a) % 1000000007;
}
int prime[1000009];
void sieve(int n) {
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (!prime[i]) {
      for (int j = 2 * i; j <= n; j++) {
        prime[j]++;
      }
    }
  }
}
bool comp(const pair<long long int, long long int> &a,
          const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
const int N = 5 * 1e5 + 9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    map<long long int, long long int> mp1;
    mp1.clear();
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      mp1[a[i]]++;
    }
    if (n == 1) {
      long long int val = a[0];
      int flag = 0;
      for (int i = 2; i <= sqrt(val); i++) {
        if (val % i == 0) {
          flag++;
          continue;
        }
      }
      if (!flag) {
        long long int r = val * val;
        cout << r << "\n";
        continue;
      } else {
        cout << "-1\n";
        continue;
      }
    }
    sort(a, a + n);
    int val = n / 2;
    map<long long int, long long int> mp;
    mp.clear();
    for (int i = 0; i < val; i++) {
      long long int w = a[i] * a[n - i - 1];
      mp[w]++;
    }
    if (mp.size() != 1) {
      cout << "-1\n";
      continue;
    }
    long long int ww = mp.begin()->first;
    if (n % 2) {
      long long int rr = a[val] * a[val];
      if (rr != ww) {
        cout << "-1\n";
        continue;
      }
    }
    int flag = 0;
    for (int i = 2; i <= sqrt(ww); i++) {
      if (ww % i == 0) {
        if (mp1.find(i) == mp1.end()) {
          flag++;
          break;
        }
        if (i * i != ww) {
          long long int kk = ww / i;
          if (mp1.find(kk) == mp1.end()) {
            flag++;
            break;
          }
        }
      }
    }
    if (flag) {
      cout << "-1\n";
    } else
      cout << ww << "\n";
  }
  return 0;
}

```