description: 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.
solution: 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--;
          }
      }
  }