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.*;
  import java.lang.*;
  import static java.lang.Math.*;
  
  public class cp{
      static BufferedReader br;
      static StringTokenizer st;
  
      public static void main(String[] args)throws IOException{
  
          br = new BufferedReader(new InputStreamReader(System.in));
          List<Long> answer = new ArrayList<Long>();
  
          for(int t=Integer.parseInt(br.readLine()); t>0; t--){
  
              st = readLine();
              int n = tokenInt();
  
              st = readLine();
              int[] arr = readIntArray(n);
  
              answer.add(new Solver().solve(n, arr));
  
          }
  
          for (long ans : answer ) System.out.println(ans);
  
      }
  
      static StringTokenizer readLine() throws IOException { return new StringTokenizer(br.readLine()); }
      static int tokenInt() { return Integer.parseInt(st.nextToken()); }
      static long tokenLong(){ return Long.parseLong(st.nextToken()); }
      static double tokenDouble(){ return Double.parseDouble(st.nextToken()); }
      static char[] tokenChar(){ return st.nextToken().toCharArray(); }
      static int[] readIntArray(int n) {
          int[] a=new int[n];
          for (int i=0; i<n; i++) a[i]=tokenInt();
          return a;
      }
  
      static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b);}
      static int lcm(int a, int b) { return (a * b) / gcd(a,b);}
  
      static void printArr(int[] arr){
          for (int i=0; i<arr.length; i++) System.out.print(arr[i] + " ");
          System.out.println();
      }
  
      static void printArr(char[] arr){
          for (int i=0; i<arr.length; i++) System.out.print(arr[i] + " ");
          System.out.println();
      }
  
      static void printArr2d(int[][] arr){
          for (int[] ar : arr) printArr(ar);
      }
  
      static void printArr2d(char[][] arr){
          for (char[] ar : arr) printArr(ar);
      }
  
  }
  
  
  class Solver{
      public long solve(int n, int[] arr) throws IOException{
  
          long prod = 0;
          if (n > 2){
              for (int i=1; i<n; i++) {
                  prod = max((long) arr[i-1] * arr[i], prod);
              }
          } else{
              prod = (long) arr[0] * arr[1];
          }
  
          return prod;
      }
  }
  
  /*
  1
  4
  11 25 37 8
  */
  