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.util.Arrays;
  import java.util.StringTokenizer;
   
  public class Test4 {
   
  	public static void main(String[] args) {
              FastScanner fs=new FastScanner();
              int T=fs.nextInt();
              PrintWriter out=new PrintWriter(System.out);
              for (int tt=0; tt<T; tt++) {
                  
              	int n=fs.nextInt();
              	int arr[] = new int[n];
              	long a = arr[0], b = arr[1];
              	long prod=1;
          		for(int i=0;i<n;i++) {
          			arr[i] = fs.nextInt();
          			
          		}
          		long min=Integer.MAX_VALUE;
          		for (int i = 0; i < n-1; i++)
          	          prod=Math.max(prod, arr[i]*arr[i+1]);
          		for (int i = n-1; i >=1; i--)
        	          prod=Math.max(prod, (long)arr[i]*(long)arr[i-1]);
          		System.out.println(prod);
              }
              out.close();
  	}
  	
  	static class FastScanner {
              BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
              StringTokenizer st=new StringTokenizer("");
              public String next() {
              while (!st.hasMoreElements())
  		try {
                      st=new StringTokenizer(br.readLine());
  		} catch (IOException e) {
                      e.printStackTrace();
  		}
  		return st.nextToken();
              }
  		
              int[] readArray(int n) {
              int[] a=new int[n]; 
              for (int i=0; i<n; i++) a[i]=nextInt();
  		return a;
              }
              int nextInt () {
                  return Integer.parseInt(next());
              }
  	}
   
  }