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.
//package Arrays_algo.TwoPointer;

import java.util.Arrays;
import java.util.Scanner;

// https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/

public class MaxProduct_K_SubsequenceSize {
    static Integer[] a;
    static int n;
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int T = scan.nextInt();
        while (T-- > 0) {
            n = scan.nextInt();
            int k = 5;
            a = new Integer[n];
            for (int i = 0; i < n; i++) a[i] = scan.nextInt();
            Arrays.sort(a);
            System.out.println(getMaxProduct(k));
        }
    }

    static long getMaxProduct(int k) {
        long ans = 1;
        if (a[n-1] == 0) {
            if (k % 2 == 1) return 0;
            else {
                for (int i = 0; i < k; i++) ans *= a[i];
            }
        } else if (a[n-1] < 0) {
            if (k % 2 == 1) {
                for (int i = n-1; i >= n-k; i--) ans *= a[i];
            } else {
                for (int i = 0; i < k; i++) ans *= a[i];
            }
        } else {
            int L = 0, R = n-1;
            if (k % 2 == 1) {
                ans *= a[n-1]; R = n-2; k--;
            }
            while (k > 0) {
                if (a[L] * a[L+1] > a[R] * a[R-1]) {
                    ans *= a[L] * a[L+1];
                    L += 2;
                } else {
                    ans *= a[R] * a[R-1];
                    R -= 2;
                }
                k -= 2;
            }
        }
        return ans;
    }

}
