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;
const long long inf = 1e18;
long long solve(multiset<int>& st, int num, int dir) {
  long long ans = 1;
  auto it = dir ? --st.end() : st.begin();
  for (int i = 1; i <= num; i++, dir ? it-- : it++) ans *= *it;
  return ans;
}
int main() {
  int t, n, x;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    long long ans = -inf;
    multiset<int> st[2];
    for (int i = 1; i <= n; i++) {
      scanf("%d", &x);
      if (!x)
        ans = 0;
      else
        st[x > 0].insert(x);
    }
    for (int i = 0; i <= 5; i++) {
      long long cur = 1;
      if ((int)st[0].size() < i || (int)st[1].size() < 5 - i) continue;
      if (i & 1)
        cur = solve(st[0], i, 1) * solve(st[1], 5 - i, 0);
      else
        cur = solve(st[0], i, 0) * solve(st[1], 5 - i, 1);
      ans = max(ans, cur);
    }
    printf("%lld\n", ans);
  }
}
