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.
/**
 * @author vivek
 * programming is thinking, not typing
 *
 * ref: https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B {

    static long pro(long A[], int n, int k)
    {
        // sorting given input array
        Arrays.sort(A);

        // variable to store final product of all element
        // of sub-sequence of size k
        long product = 1;

        // CASE I
        // If max element is 0 and
        // k is odd then max product will be 0
        if (A[n - 1] == 0 && k % 2 != 0)
            return 0;

        // CASE II
        // If all elements are negative and
        // k is odd then max product will be
        // product of rightmost-subarray of size k
        if (A[n - 1] <= 0 && k % 2 != 0) {
            for (int i = n - 1; i >= n - k; i--)
                product *= A[i];
            return product;
        }

        // else
        // i is current left pointer index
        int i = 0;

        // j is current right pointer index
        int j = n - 1;

        // CASE III
        // if k is odd and rightmost element in
        // sorted array is positive then it
        // must come in subsequence
        // Multiplying A[j] with product and
        // correspondingly changing j
        if (k % 2 != 0) {
            product *= A[j];
            j--;
            k--;
        }

        // CASE IV
        // Now k is even
        // Now we deal with pairs
        // Each time a pair is multiplied to product
        // ie.. two elements are added to subsequence each time
        // Effectively k becomes half
        // Hence, k >>= 1 means k /= 2
        k >>= 1;

        // Now finding k corresponding pairs
        // to get maximum possible value of product
        for (int itr = 0; itr < k; itr++) {
            // product from left pointers
            long left_product = A[i] * A[i + 1];

            // product from right pointers
            long right_product = A[j] * A[j - 1];

            // Taking the max product from two choices
            // Correspondingly changing the pointer's position
            if (left_product > right_product) {
                product *= left_product;
                i += 2;
            }
            else {
                product *= right_product;
                j -= 2;
            }
        }

        // Finally return product
        return product;
    }

    private static void solveTC(int __) {

        /* For Google */
//		ans.append("Case #").append(__).append(": ");

        //code start

        int n=scn.nextInt();
        long[] arr=scn.nextLongArray(n);

        print(pro(arr,n,5));

        //code end

        print("\n");
    }

    /**
     * Don't copy other's templates, make your own <br>
     * It would be much more beneficial
     */

    public static void main(String[] args) {
        scn = new Scanner();
        ans = new StringBuilder();
        int t = scn.nextInt();
//		int t = 1;

//        int limit=               ;
//        sieve(limit);

        /*
        try {
            System.setOut(new PrintStream(new File("file_i_o\\output.txt")));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
*/
        for (int i = 1; i <= t; i++) {
            solveTC(i);
        }
        System.out.print(ans);
    }
    //Stuff for prime start

    /**
     * List containing prime numbers <br>
     * <b>i<sup>th</sup></b> position contains <b>i<sup>th</sup></b> prime number <br>
     * 0th index is <b>null</b>
     */
    private static ArrayList<Integer> listOfPrimes;

    /**
     * query <b>i<sup>th</sup></b> element to get if its prime of not
     */
    private static boolean[] isPrime;

    /**
     * Performs Sieve of Erathosnesis and initialise isPrime array and listOfPrimes list
     *
     * @param limit the number till which sieve is to be performed
     */
    private static void sieve(int limit) {
        listOfPrimes = new ArrayList<>();
        listOfPrimes.add(null);
        boolean[] array = new boolean[limit + 1];

        Arrays.fill(array, true);
        array[0] = false;
        array[1] = false;

        for (int i = 2; i <= limit; i++) {
            if (array[i]) {
                for (int j = i * i; j <= limit; j += i) {
                    array[j] = false;
                }
            }
        }
        isPrime = array;
        for (int i = 0; i <= limit; i++) {
            if (array[i]) {
                listOfPrimes.add(i);
            }
        }
    }
    //stuff for prime end

    /**
     * Calculates the Least Common Multiple of two numbers
     *
     * @param a First number
     * @param b Second Number
     * @return Least Common Multiple of <b>a</b> and <b>b</b>
     */
    private static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    /**
     * Calculates the Greatest Common Divisor of two numbers
     *
     * @param a First number
     * @param b Second Number
     * @return Greatest Common Divisor of <b>a</b> and <b>b</b>
     */
    private static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void print(Object obj) {
        ans.append(obj.toString());
    }


    static Scanner scn;
    static StringBuilder ans;

    //Fast Scanner
    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        public Scanner() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
/*
            try {
                br = new BufferedReader(new
                        InputStreamReader(new FileInputStream(new File("file_i_o\\input.txt"))));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
*/
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] nextIntArray(int n) {
            int[] array = new int[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        Integer[] nextIntegerArray(int n) {
            Integer[] array = new Integer[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextInt();
            }
            return array;
        }

        long[] nextLongArray(int n) {
            long[] array = new long[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextLong();
            }
            return array;
        }

        String[] nextStringArray() {
            return nextLine().split(" ");
        }

        String[] nextStringArray(int n) {
            String[] array = new String[n];
            for (int i = 0; i < n; i++) {
                array[i] = next();
            }
            return array;
        }
    }
}
