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.io.*;
  import java.util.*;
   
  public class Cherry {
  
      public static void main(String[] args) throws IOException {
          FastScanner in = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int t = in.nextInt();
          for (int i = 1; i <= t; i++)  {
          	int n = in.nextInt();
          	long[] val = new long[n];
          	for (int k = 0; k < n; k++)  {
          		val[k] = in.nextLong();
          	}
          	
          	if (n == 2)  {
          		out.println(val[0] * val[1]);
          	}
          	else  {
              	long maxproduct = 0;
              	for (int k = 1; k < n - 1; k++)  {
              		if ((val[k] >= val[k - 1]) && (val[k] >= val[k + 1]))  {
              			long temp1 = val[k] * val[k - 1];
              			long temp2 = val[k] * val[k + 1];
              			
              			if (Math.max(temp1, temp2) > maxproduct)  {
              				maxproduct = Math.max(temp1, temp2);
              			}
              		}
              	}
              	
      			maxproduct = Math.max(maxproduct, val[0] * val[1]);
      			maxproduct = Math.max(maxproduct, val[n - 2] * val[n - 1]);
              	
              	out.println(maxproduct);
          	}
          }
          out.close();
      }
      
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try { 
                      st = new StringTokenizer(br.readLine());                               
                  } catch (IOException e) {
                      // noop
                  }
              return st.nextToken();
          }
          
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
          double nextDouble()  {
              return Double.parseDouble(next());
          }
      }
  
  }