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.*;
import java.io.*;
public class Solution{
    public static void main(String arg[]){
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t > 0)
        {
		    int i, j, n = scan.nextInt();
			long p, q, r;
			long [] arr = new long[n]; 
			LinkedList <Long> list1 = new LinkedList<Long>();
			LinkedList <Long> list2 = new LinkedList<Long>();
			for (i = 0; i < n; ++i)
			{
			arr[i] = scan.nextLong();
			}
			i = 0;
			j = 1;
			p = arr[0];
			q = arr[0];
			r = 0;
			list1.add(arr[0]);
			list2.add(arr[0]);
			while (j < n)
			{
			while (list1.size() > 0 && list1.getLast() < arr[j])
			{
			list1.removeLast();
			}
			list1.add(arr[j]);
			while (list2.size() > 0 && list2.getLast() > arr[j])
			{
			list2.removeLast();
			}
			list2.add(arr[j]);
			j++;
			r = Math.max(r, list1.getFirst()*list2.getFirst());
			while (i + 1 < j && arr[i] != list2.getFirst())
			{
			if (arr[i] == list1.getFirst())
			{
			list1.removeFirst();
			}
			i++;
			}
			if (i + 1 < j)
			{
			if (arr[i] == list1.getFirst())
			{
			list1.removeFirst();
			}
			list2.removeFirst();
			i++;
			}
			if (i + 1 < j)
			{
			r = Math.max(r, list1.getFirst()*list2.getFirst());
			}
			}
			if (i + 1 < j)
			{
			r = Math.max(r, list1.getFirst()*list2.getFirst());
			}
			System.out.println(r);
            t--;
        }
    }
}