You are given an array of integers a_1,a_2,…,a_n. Find the maximum possible value of a_ia_ja_ka_la_t among all five indices (i, j, k, l, t) (i<j<k<l<t).

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (5≤ n≤ 10^5) — the size of the array.

The second line of each test case contains n integers a_1,a_2,…,a_n (-3× 10^3≤ a_i≤ 3× 10^3) — given array.

It's guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

Output

For each test case, print one integer — the answer to the problem.

Example

Input


4
5
-1 -2 -3 -4 -5
6
-1 -2 -3 1 2 -1
6
-1 0 0 0 -1 -1
6
-9 -7 -5 -3 -2 1


Output


-120
12
0
945

Note

In the first test case, choosing a_1,a_2,a_3,a_4,a_5 is a best choice: (-1)⋅ (-2) ⋅ (-3)⋅ (-4)⋅ (-5)=-120.

In the second test case, choosing a_1,a_2,a_3,a_5,a_6 is a best choice: (-1)⋅ (-2) ⋅ (-3)⋅ 2⋅ (-1)=12.

In the third test case, choosing a_1,a_2,a_3,a_4,a_5 is a best choice: (-1)⋅ 0⋅ 0⋅ 0⋅ (-1)=0.

In the fourth test case, choosing a_1,a_2,a_3,a_4,a_6 is a best choice: (-9)⋅ (-7) ⋅ (-5)⋅ (-3)⋅ 1=945.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll powerflow(ll x, ll n) {
  ll res;
  if (n == 0) {
    return 1;
  }
  res = powerflow(x, n - 1);
  if (INT_MAX / res < x) {
    return -1;
  }
  return x * res;
}
struct compare {
  bool operator()(const int& l, const int& r) { return l > r; }
};
void solve() {
  ll n;
  cin >> n;
  vector<ll> v(n);
  for (int i = 0; i < (n); ++i) cin >> v[i];
  sort(v.begin(), v.end());
  ll ans = 1;
  for (int i = 0; i < (5); ++i) ans *= v[i];
  ans = max(v[0] * v[1] * v[n - 1] * v[n - 2] * v[n - 3], ans);
  ans = max(v[0] * v[1] * v[2] * v[3] * v[n - 1], ans);
  ans = max(v[n - 1] * v[n - 2] * v[n - 3] * v[n - 4] * v[n - 5], ans);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
