#include <bits/stdc++.h>
const int mod = 1e9 + 7;
using namespace std;
template <typename T>
class fenwick_tree {
  vector<T> t;
  vector<T> arr;
  string convert_to_binary(T num) {
    string s = "";
    for (; num; num >>= 1) {
      s.push_back((num & 1) + '0');
    }
    s.push_back('0');
    reverse(s.begin(), s.end());
    return s;
  }
  void build() {
    for (size_t i = 0; i < arr.size(); ++i) {
      inc(i, arr[i]);
    }
  }

 public:
  fenwick_tree(vector<T>& arr) {
    this->arr = arr;
    t.resize(arr.size());
    build();
  }
  fenwick_tree(vector<T>&& arr) {
    this->arr = move(arr);
    t.resize(this->arr.size());
    build();
  }
  void inc(int i, T val) {
    for (; i < t.size(); i = i | (i + 1)) {
      t[i] += val;
    }
  }
  T get(int i) {
    T num = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) {
      num += t[i];
    }
    return num;
  }
  T sum(int i, int j) { return get(j) - get(i - 1); }
};
template <typename T>
T gcd(T a, T b) {
  if (a < b) {
    swap(a, b);
  }
  if (!b) {
    return a;
  }
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    for (auto& i : arr) {
      cin >> i;
    }
    sort(arr.begin(), arr.end());
    long long i = 0, j = n - 1, f = arr[i] * arr[j];
    i++;
    j--;
    bool flg = true;
    for (; i <= j; ++i, --j) {
      if (arr[i] * arr[j] != f) {
        flg = false;
        break;
      }
    }
    int cnt = 0;
    for (long long i = 2; i * i <= f; ++i) {
      if (f % i == 0) {
        cnt += (f / i != i) + 1;
      }
    }
    flg &= (cnt == n);
    cout << ((flg) ? f : -1) << endl;
  }
  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>
const int mod = 1e9 + 7;
using namespace std;
template <typename T>
class fenwick_tree {
  vector<T> t;
  vector<T> arr;
  string convert_to_binary(T num) {
    string s = "";
    for (; num; num >>= 1) {
      s.push_back((num & 1) + '0');
    }
    s.push_back('0');
    reverse(s.begin(), s.end());
    return s;
  }
  void build() {
    for (size_t i = 0; i < arr.size(); ++i) {
      inc(i, arr[i]);
    }
  }

 public:
  fenwick_tree(vector<T>& arr) {
    this->arr = arr;
    t.resize(arr.size());
    build();
  }
  fenwick_tree(vector<T>&& arr) {
    this->arr = move(arr);
    t.resize(this->arr.size());
    build();
  }
  void inc(int i, T val) {
    for (; i < t.size(); i = i | (i + 1)) {
      t[i] += val;
    }
  }
  T get(int i) {
    T num = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) {
      num += t[i];
    }
    return num;
  }
  T sum(int i, int j) { return get(j) - get(i - 1); }
};
template <typename T>
T gcd(T a, T b) {
  if (a < b) {
    swap(a, b);
  }
  if (!b) {
    return a;
  }
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    for (auto& i : arr) {
      cin >> i;
    }
    sort(arr.begin(), arr.end());
    long long i = 0, j = n - 1, f = arr[i] * arr[j];
    i++;
    j--;
    bool flg = true;
    for (; i <= j; ++i, --j) {
      if (arr[i] * arr[j] != f) {
        flg = false;
        break;
      }
    }
    int cnt = 0;
    for (long long i = 2; i * i <= f; ++i) {
      if (f % i == 0) {
        cnt += (f / i != i) + 1;
      }
    }
    flg &= (cnt == n);
    cout << ((flg) ? f : -1) << endl;
  }
  return 0;
}

```