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 P07 {
  	static class FastReader{
  		BufferedReader br ;
  		StringTokenizer st;
  		public FastReader() {
  			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());
  		}
  		
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			}catch(IOException e) {
  				e.printStackTrace();
  			}
  			return str ;
  		}
  	}
  	
   	public static  void main(String[] args)throws IOException {
  		FastReader fs = new FastReader();
  		int t = fs.nextInt();
  		while(t-->0) {
  			int n = fs.nextInt();
  			long a[]= new long [n];
  			long max =Long.MIN_VALUE;
  			for(int i =0;i<n;i++) {
  				a[i]= fs.nextLong();
  			}
  			for(int i =0;i<n-1;i++) {
  				max=Math.max(max,a[i]*a[i+1]);
  			}
  			for(int i =n-1;i>0;i--) {
  				max=Math.max(max,a[i]*a[i-1]);
  			}
  			System.out.println(max);
  		}
  		
      }
  	private static int gcd(int i, int j) {
  		if(j==0) {
  			return i;
  		}
  		return gcd(j,i%j);
  	}
  	
  }
  