#include <bits/stdc++.h>
using namespace std;
int long long a[305], b[305];
vector<int> v, v2;
vector<int long long> prime;
char sieve[10000007];
void primeSieve(int long long n) {
  sieve[0] = sieve[1] = 1;
  prime.push_back(2);
  for (int long long i = 4; i <= n; i += 2) sieve[i] = 1;
  int long long sqrtn = sqrt(n);
  for (int long long i = 3; i <= sqrtn; i += 2) {
    if (sieve[i] == 0) {
      for (int long long j = i * i; j <= n; j += 2 * i) sieve[j] = 1;
    }
  }
  for (int i = 3; i <= n; i += 2)
    if (sieve[i] == 0) prime.push_back(i);
}
int long long NOD(int long long n) {
  int long long sqrtn = sqrt(n);
  int long long res = 1;
  for (int long long i = 0; i < prime.size() && prime[i] <= sqrtn; i++) {
    if (n % prime[i] == 0) {
      int long long p = 0;
      while (n % prime[i] == 0) {
        n /= prime[i];
        p++;
      }
      sqrtn = sqrt(n);
      p++;
      res *= p;
    }
  }
  if (n != 1) {
    res *= 2;
  }
  return res;
}
int main() {
  int t, n, kase = 0;
  scanf("%d", &t);
  primeSieve(10000000);
  while (t--) {
    scanf("%d", &n);
    int long long ans = 1;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
    }
    if (n == 1) {
      if (sieve[a[0]] == 1)
        printf("-1\n");
      else
        printf("%lld\n", a[0] * a[0]);
      continue;
    }
    sort(a, a + n);
    if (sieve[a[0]] == 1) {
      printf("-1\n");
      continue;
    }
    ans = a[0] * a[n - 1];
    if (NOD(ans) - 2 != n) {
      printf("-1\n");
      continue;
    }
    int flag = 0;
    int j = n - 1;
    for (int i = 0; i <= n / 2; i++) {
      if (a[i] * a[j] != ans) {
        flag = 1;
        break;
      }
      j--;
    }
    if (flag == 1 || ans < 0)
      printf("-1\n");
    else
      printf("%lld\n", ans);
  }
  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;
int long long a[305], b[305];
vector<int> v, v2;
vector<int long long> prime;
char sieve[10000007];
void primeSieve(int long long n) {
  sieve[0] = sieve[1] = 1;
  prime.push_back(2);
  for (int long long i = 4; i <= n; i += 2) sieve[i] = 1;
  int long long sqrtn = sqrt(n);
  for (int long long i = 3; i <= sqrtn; i += 2) {
    if (sieve[i] == 0) {
      for (int long long j = i * i; j <= n; j += 2 * i) sieve[j] = 1;
    }
  }
  for (int i = 3; i <= n; i += 2)
    if (sieve[i] == 0) prime.push_back(i);
}
int long long NOD(int long long n) {
  int long long sqrtn = sqrt(n);
  int long long res = 1;
  for (int long long i = 0; i < prime.size() && prime[i] <= sqrtn; i++) {
    if (n % prime[i] == 0) {
      int long long p = 0;
      while (n % prime[i] == 0) {
        n /= prime[i];
        p++;
      }
      sqrtn = sqrt(n);
      p++;
      res *= p;
    }
  }
  if (n != 1) {
    res *= 2;
  }
  return res;
}
int main() {
  int t, n, kase = 0;
  scanf("%d", &t);
  primeSieve(10000000);
  while (t--) {
    scanf("%d", &n);
    int long long ans = 1;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
    }
    if (n == 1) {
      if (sieve[a[0]] == 1)
        printf("-1\n");
      else
        printf("%lld\n", a[0] * a[0]);
      continue;
    }
    sort(a, a + n);
    if (sieve[a[0]] == 1) {
      printf("-1\n");
      continue;
    }
    ans = a[0] * a[n - 1];
    if (NOD(ans) - 2 != n) {
      printf("-1\n");
      continue;
    }
    int flag = 0;
    int j = n - 1;
    for (int i = 0; i <= n / 2; i++) {
      if (a[i] * a[j] != ans) {
        flag = 1;
        break;
      }
      j--;
    }
    if (flag == 1 || ans < 0)
      printf("-1\n");
    else
      printf("%lld\n", ans);
  }
  return 0;
}

```