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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.*;
  import java.math.*;
  import static java.lang.Math.min;
  import static java.lang.Math.max;
  import static java.lang.Math.abs;
  
  /* https://codeforces.com/contest/1554/problem/A*/
  public class A {
      static Node st[];
      static FastScanner fs = new FastScanner();
      static PrintWriter fop = new PrintWriter(System.out);
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) {
          try {
              solve();
          } catch (Exception e) {
              return;
          }
          // solve();
      }
  
      static void solve() throws IOException {
  
          int T = fs.nextInt();
          while (T-- > 0) {
              long ans = 0,nans = 0;
              int n = fs.nextInt();
              int a[] = fs.readArray(n);
              for (int i = 0; i < n - 1; i++) {
                  nans = (long)a[i] * a[i + 1];
                  ans = nans>ans?nans:ans;
              }
              fop.println(ans);
          }
          fop.flush();
          fop.close();
  
      }
  
      static class Node {
          int min;
          int max;
      }
  
      static Node MaxMinUtil(Node st[], int ss, int se, int qs, int qe, int index) {
          Node temp = new Node();
          Node left, right;
          if (qs <= ss && qe >= se)
              return st[index];
          if (se < qs || ss > qe) {
              temp.min = Integer.MAX_VALUE;
              temp.max = Integer.MIN_VALUE;
              return temp;
          }
          int mid = ss + (ss + se) / 2;
          left = MaxMinUtil(st, ss, mid, qs, qe, 2 * index + 1);
          right = MaxMinUtil(st, mid + 1, se, qs, qe, 2 * index + 2);
          temp.min = min(left.min, right.min);
          temp.max = max(left.max, right.max);
          return temp;
      }
  
      static Node MaxMin(Node[] st, int n, int qs, int qe) {
          return MaxMinUtil(st, 0, n - 1, qs, qe, 0);
      }
  
      static void constructSTUtil2(int[] arr, int ss, int se, Node[] st, int si) {
          if (ss == se) {
              st[si].min = arr[ss];
              st[si].max = arr[ss];
              return;
          }
  
          int mid = ss + (ss + se) / 2;
          constructSTUtil2(arr, ss, mid, st, si * 2 + 1);
          constructSTUtil2(arr, mid + 1, se, st, si * 2 + 2);
          st[si].min = min(st[si * 2 + 1].min, st[si * 2 + 2].min);
          st[si].max = min(st[si * 2 + 1].max, st[si * 2 + 2].max);
          // return st[si];
      }
  
      static void constructST(int arr[], int n) {
          int h = (int) (Math.ceil(Math.log(n) / Math.log(2)));
  
          int max_size = 2 * (int) Math.pow(2, h) - 1;
          st = new Node[max_size];
          constructSTUtil2(arr, 0, n - 1, st, 0);
          // return st;
      }
      // above function for both ^^^
  
      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) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
          long[] readLongArray(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextLong();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }
  