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.*;
   
  public class SolutionA{
         public static void main(String[] args) throws Exception{
                 Fast sc=new Fast();
        PrintWriter out=new PrintWriter(System.out);
         int t=sc.nextInt();
         while(t-->0)
         {
              int n=sc.nextInt();
              long ans=0;
              long ti=0;
              long[] ar=new long[n];
              for(int i=0;i<n;i++){
                ar[i]=sc.nextLong();
              }
              long max=ar[0];
              long min=ar[0];
              for(int i=1;i<n;i++){
                max=Math.max(max,ar[i]);
                min=Math.min(min,ar[i]);
                ans=Math.max(ans,min*max);
                ans=Math.max(ans,ar[i]*ar[i-1]);
              }
              out.println(ans);
                    
         }
      out.close();
  }
  
  static void Sort(int[] ar){
      ArrayList<Integer> al=new ArrayList<>();
      for(int i=0;i<ar.length;i++){al.add(ar[i]);}
      Collections.sort(al);
      for(int i=0;i<al.size();i++)ar[i]=al.get(i);
    }
  static long gcd(long a,long b){
      if(b==0) return a;
      else return  gcd(b,a%b);
  }
  }
  class Pair{
      int x, y;
       Pair(int x,int y){
          this.x=x;
          this.y=y;
       }
  }
  class Fast{
    BufferedReader br; 
    StringTokenizer st;
    public Fast(){ br=new BufferedReader(new InputStreamReader(System.in)); }
    String next() 
          {  while (st == null || !st.hasMoreElements()) 
              {   try
                  {     st = new StringTokenizer(br.readLine()); } 
                  catch (IOException  e) 
                  {    e.printStackTrace(); } 
              } 
              return st.nextToken(); 
          }
    int nextInt(){  return Integer.parseInt(next()); }
    long nextLong(){     return Long.parseLong(next()); } 
    double nextDouble() {     return Double.parseDouble(next()); } 
    String nextLine() 
          { 
              String str = ""; 
              try{      str = br.readLine(); } 
              catch (IOException e)  {  e.printStackTrace(); } 
              return str; 
          } 
  }
      
  