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.BufferedInputStream;
  import java.util.Scanner;
  
  /**
   * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
   *
   * To modify the template, go to Preferences -> Editor -> File and Code Templates -> Other
   */
  
  public class Main {
     public static void main(String[] args) {
        // Write your solution here
        Scanner scanner = new Scanner(new BufferedInputStream(System.in));
        int[] a_s = new int[100005];
        int t = scanner.nextInt();
  
        while (t-- > 0) {
           int n = scanner.nextInt();
  
           for (int i = 0; i < n; i++) {
              a_s[i] = scanner.nextInt();
           }
  
           int l, h;
  
           long max = 0;
  
           if (a_s[0] < a_s[1]) {
              l = 0; h = 1;
           } else {
              l = 1; h = 0;
           }
  
           max = (long) a_s[l] * a_s[h];
  
           for (int i = 2; i < n; i++) {
              if (a_s[i] <= a_s[l]) {
                 if (i + 1 >= n) {
                    break;
                 } else {
                    if (a_s[i] < a_s[i + 1]) {
                       l = i; h = i + 1;
                    } else {
                       l = i + 1; h = i;
                    }
                    i++;
                    max = (long) a_s[l] * a_s[h] > max ? (long) a_s[l] * a_s[h] : max;
                 }
              } else if (a_s[i] >= a_s[h]) {
                 if (l < h) {
                    l = h;
                 }
                 h = i;
                 max = (long) a_s[l] * a_s[h] > max ? (long) a_s[l] * a_s[h] : max;
              } else {
                 if (l < h) {
                    l = i;
                    max = (long) a_s[l] * a_s[h] > max ? (long) a_s[l] * a_s[h] : max;
                 } else {
                    if (i + 1 >= n) {
                       break;
                    } else {
                       if (a_s[i] < a_s[i + 1]) {
                          l = i; h = i + 1;
                       } else {
                          l = i + 1; h = i;
                       }
                       i++;
                       max = (long) a_s[l] * a_s[h] > max ? (long) a_s[l] * a_s[h] : max;
                    }
  //                  h = i;
  //                  max = (long) a_s[l] * a_s[h] > max ? (long) a_s[l] * a_s[h] : max;
                 }
              }
           }
  
           System.out.println(max);
        }
     }
  }