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.
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {
	public static void main (String[] args) {
	   // Scanner scan = new Scanner(System.in);
    	FastScanner scan = new FastScanner();
        int t = scan.nextInt();
        while(t-- > 0) {
            int n = scan.nextInt();
            List<Long> ls = new ArrayList<>();
            for(int i = 0; i < n; i++) {
                long num = scan.nextLong();
                // if(num != 0) {
                    ls.add(num);
                // }
            }
            
            // if(ls.size() < 5) {
                // System.out.println(0);
            // } else {
                Collections.sort(ls);
                long max = Long.MIN_VALUE;
                n = ls.size();
                long firstFour = ls.get(0) * ls.get(1) * ls.get(2) * ls.get(3);
                long lastFour = ls.get(n - 4) * ls.get(n - 1) * ls.get(n - 2) * ls.get(n - 3);
                long lastFive = ls.get(n - 1) * ls.get(n - 2) * ls.get(n - 3) * ls.get(n - 4) * ls.get(n - 5);
                max = Math.max(max, firstFour * ls.get(n - 1));
                max = Math.max(max, lastFour * ls.get(0));
                max = Math.max(max, lastFive);
                max = Math.max(max, firstFour * ls.get(4)); // firstFive
                max = Math.max(max, ls.get(0) * ls.get(1) * ls.get(n - 1) * ls.get(n - 2) * ls.get(n - 3));
                System.out.println(max);
            // }
        }
    }
/*
    int n = scan.nextInt();
            int[] arr = scan.readArray(n);
    
    int n = scan.nextInt();
            int m = scan.nextInt();
            int[][] arr = new int[n][m];
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < m; j++) {
                    arr[i][j] = scan.nextInt();
                }
            }
    
*/
    
    
    static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
    
}