You are given n integers a_1, a_2, …, a_n. Find the maximum value of max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r) over all pairs (l, r) of integers for which 1 ≤ l < r ≤ n.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print a single integer — the maximum possible value of the product from the statement.

Example

Input


4
3
2 4 3
4
3 2 3 1
2
69 69
6
719313 273225 402638 473783 804745 323328


Output


12
6
4761
381274500335

Note

Let f(l, r) = max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r).

In the first test case, 

  * f(1, 2) = max(a_1, a_2) ⋅ min(a_1, a_2) = max(2, 4) ⋅ min(2, 4) = 4 ⋅ 2 = 8. 
  * f(1, 3) = max(a_1, a_2, a_3) ⋅ min(a_1, a_2, a_3) = max(2, 4, 3) ⋅ min(2, 4, 3) = 4 ⋅ 2 = 8. 
  * f(2, 3) = max(a_2, a_3) ⋅ min(a_2, a_3) = max(4, 3) ⋅ min(4, 3) = 4 ⋅ 3 = 12. 



So the maximum is f(2, 3) = 12.

In the second test case, the maximum is f(1, 2) = f(1, 3) = f(2, 3) = 6.
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.Collections;

public class main {
    public static void main(String[] args) {
        doCalculation();
    }

    public static void doCalculation () {
        Scanner sc = new Scanner(System.in);

        long n = sc.nextLong();

        for (int i = 0; i < n; i++) {
            int numbersInRow = sc.nextInt();
            long [] numbersToRead = new long[numbersInRow];
            for(int j = 0; j < numbersInRow; j++) {
                numbersToRead[j] = sc.nextLong();
            }

            long maxMultiplication = 0;

            for(int k = 0; k < numbersInRow - 1; k++) {
                final long product = numbersToRead[k] * numbersToRead[k+1];
                if (product > maxMultiplication) {
                    maxMultiplication = product;
                }
            }


            /*for(int k = 0; k < numbersInRow; k++) {
                for (int m = k + 1; m < numbersInRow; m++) {
                    //Von jetzt an von 1 bis 2; von 1 bis 3; von 2 bis 3
                    long[] subarray = subArray(numbersToRead, k, m);
                    //System.out.println(Arrays.toString(subarray));

                    long min = Arrays.stream(subarray).min().getAsLong();
                    //System.out.println(min);

                    long max = Arrays.stream(subarray).max().getAsLong();
                    //System.out.println(max);

                    long multiplicationEveryArray = min * max;

                    if(multiplicationEveryArray > maxMultiplication) {
                        maxMultiplication = multiplicationEveryArray;
                    }
                }
            } */
            System.out.println(maxMultiplication);
        }
    }
    public static long[] subArray(long[] array, int beg, int end) {
        return Arrays.copyOfRange(array, beg, end + 1);
    }
}