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.util.StringTokenizer;
  
  public class Main {
  	public static int N;
  
  	public static void main(String[] args) throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st;
  		
  		int T = Integer.parseInt(br.readLine());
  		
  		for(int testcase = 0; testcase < T; testcase++) {
  			N = Integer.parseInt(br.readLine());
  			st = new StringTokenizer(br.readLine());
  			long[] arr = new long[N + 2];
  			
  			for(int i = 1; i <= N; i++) {
  				arr[i] = Long.parseLong(st.nextToken());
  			}
  			arr[N + 1] = arr[N] - 1;
  			
  			System.out.println(solve(arr));
  		}
  
  	}
  	
  	public static long solve(long[] arr) {
  		long max = 0;
  		
  		for(int i = 1; i <= N; i++) {
  			if(i == 1) {
  				if(arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) {
  					max = Math.max(max, arr[i] * arr[i + 1]);
  				}
  			}
  			else if(i == N) {
  				if(arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) {
  					max = Math.max(max, arr[i - 1] * arr[i]);
  				}
  			}
  			else {
  				if(arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) {
  					max = Math.max(max, arr[i - 1] * arr[i]);
  					max = Math.max(max, arr[i] * arr[i + 1]);
  				}
  			}
  		}
  		
  		return max;
  	}
  
  }
  