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.Scanner;
  
  public class Codeforces {
  	static Scanner sc=new Scanner(System.in);
  	
  	public static void main(String[] args) {
  		int t=1;
  		t=sc.nextInt();
  		
  		while(t-->0) {
  			solve();
  		}
  		
  	}
  
  	private static void solve() {
  		
  		int size=sc.nextInt();
  		long arr[]=new long[size];
  		for(int i=0;i<size;i++) {
  			arr[i]=sc.nextLong();
  		}
  		if(size==2) {
  			System.out.println(arr[0]*arr[1]);
  			return;
  		}
  		long max=0;
  		for(int i=0;i<size-1;i++) {
  			if(arr[i]*arr[i+1]>max) {
  				max=arr[i]*arr[i+1];
  			}
  		}
  		
  		System.out.println(max);
  		
  		
  		/*
  		 * long n=sc.nextLong(); if(n<6) { System.out.println(15); return ; } long
  		 * small=n/6;
  		 * 
  		 * long left=n%6; if(left==0) { System.out.println(small*15); }else if(left<=2)
  		 * { long time=(small-1)*15+20; System.out.println(time); }else if(left<=4) {
  		 * long time=(small-1)*15+25; System.out.println(time); }else {
  		 * System.out.println((small+1)*15); }
  		 */
  		
  		/*
  		 * int x=sc.nextInt(); System.out.println((x+1)/10);
  		 */		
  		/*int n=sc.nextInt();
  		int arr[]=new int[n];
  		for(int i=0;i<n;i++) {
  			arr[i]=sc.nextInt();
  		}
  		
  		int sl=Integer.MIN_VALUE;
  		int fl=Integer.MIN_VALUE;
  		
  		for(int i=0;i<n;i++) {
  			if(arr[i]>fl) {
  				sl=fl;
  				fl=arr[i];
  			}else if(arr[i]>sl && arr[i]!=fl) {
  				sl=arr[i];
  			}
  		}
  		System.out.println(fl+"  "+sl);
  		if(sl==Integer.MIN_VALUE) {
  			System.out.println(fl*fl);
  		}else {
  			System.out.println((long)fl*sl);
  		}
  		
  		*/
  	}
  
  	/*private static void solve() {
  		int l=sc.nextInt();
  		int t=sc.nextInt();
  		String str=sc.next();//.substring(0,l);
  		while(t-->0) {
  			int low=sc.nextInt();
  			int upper=sc.nextInt();
  			if(low<0 || upper>l) {
  				return ;
  			}
  			String subStr=str.substring(low-1,upper);
  			
  			int count=0;
  			for(int i=0;i<subStr.length();i++) {
  				count+= subStr.charAt(i)-'a'+1;
  			}
  			System.out.println(count);
  			
  			
  		}
  	}*/
  }
  