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.Scanner;

public class cherry
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int testNo = sc.nextInt();

        while(testNo-->0)
        {
            int n = sc.nextInt();
//            int[] array = new int[n+1];
//            array[0]=0;

            int min = Integer.MAX_VALUE;
            int max = Integer.MIN_VALUE;
            long res = 1;
            long res1=1;
            //int temp1=1;
            int another=1;

            for(int i=0;i<2;i++)
            {
                another = sc.nextInt();
                min = Math.min(min,another);
                max = Math.max(max,another);
                res1*=another;
            }


            for(int i=3;i<n;i++)
            {
                 int temp =sc.nextInt();
                min = Math.min(min,temp);
                max = Math.max(max,temp);
                long mediate = max*min;

                res = Math.max(mediate, (long) temp*another);
                another=temp;
            }

            res = Math.max(res,res1);

            System.out.println(res);
        }

    }
}
