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.
import java.util.Scanner;
import java.util.Arrays;
public class Main {
	public static void main(String[]args){
		Scanner sc = new Scanner(System.in);
	
		int l = sc.nextInt();
			
		for(int x = 0 ; x< l ; x++) {
			long max =0;
			long min =0;
			long mult = 1;
			long multl = 1;
		   
    		int n = sc.nextInt();
    			
    		int a[] = new int[(int) n];
    		for(int i=0; i<n ; i++) {
    			a[i] = sc.nextInt();
    		}
    			
    			
    		for(int i=0; i<n ; i++){
    			if(a[i] > max) {
    				max = a[i];
    			}
    		}
    			
    	   long xyz = 0;
    		   
    		for(int i=0 ; i<n ; i++) {
    			if(a[i] == max) {
    				if(i==0 ) {
    					mult = max * a[(i+1)];
    					xyz =  Math.max(xyz,  mult);
    						
    				}else if(i == n-1) {
    				    mult = max * a[(i-1)];
    					xyz =  Math.max(xyz,  mult);
    				}else {
    					mult = max * a[(i+1)];
    				    multl = max * a[(i-1)];
    					if(mult >=multl) {
    					    xyz = Math.max(xyz ,mult);
    					}else {
    					   	xyz = Math.max(xyz ,multl);
    				    	//System.out.println(multl);
    				    }
    				}
    			}
    		}
    		
    		System.out.println(xyz);
		}	
	}
}