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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.nio.Buffer;
  import java.util.StringTokenizer;
  
  public class A {
      static StringTokenizer st;
      static BufferedReader br;
      static PrintWriter pw;
  
      static String nextToken() {
          try {
              while (st == null || !st.hasMoreTokens()) {
                  st = new StringTokenizer(br.readLine());
              }
          } catch (IOException e) {
              e.printStackTrace();
          }
          return st.nextToken();
      }
  
      static int nextInt() {
          return Integer.parseInt(nextToken());
      }
  
      static long nextLong() {
          return Long.parseLong(nextToken());
      }
  
      static double nextDouble() {
          return Double.parseDouble(nextToken());
      }
  
      public static void main(String[] args) {
          br = new BufferedReader(new InputStreamReader(System.in));
          pw = new PrintWriter(System.out);
          run();
          pw.close();
      }
  
      private static void run() {
          int t = nextInt();
          for (int i = 0; i < t; i++) {
              int n = nextInt();
              long[] a = new long[n];
              for (int j = 0; j < n; j++) {
                  a[j] = nextInt();
              }
              long ans = 0;
              for (int j = 1; j < n; j++) {
                  ans = Math.max(ans, a[j] * a[j - 1]);
              }
              pw.println(ans);
  
          }
      }
  }
  