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.*;
import java.io.*;
public class A_Cherry{
    public static void main(String[] args) throws IOException{
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      int t=Integer.parseInt(br.readLine());
      while(t>0){
        int n=Integer.parseInt(br.readLine());
        long ar[]=new long[n];

        String []s=br.readLine().split(" ");
        long max=0;
        long maxvalu=0;
        for(int i=0;i<n;i++){ar[i]=Long.parseLong(s[i]); if(max<ar[i]) max=ar[i];}
            ArrayList<Integer> al=new ArrayList<Integer>();
            for(int i = 0; i < n; i++) if(ar[i]==max) al.add(i);
            for(int e:al){

                if(e==0){
                 long k=ar[e]*ar[e+1];
                 if(maxvalu<k) maxvalu=k;
                }
                else if(e==n-1){
                    long k=ar[e]*ar[e-1];
                 if(maxvalu<k) maxvalu=k;
                
                }
                else{
                    long k=ar[e]*ar[e+1];
                    long l=ar[e]*ar[e-1];
                    long m=Math.max(k, l);
                    if(m>maxvalu) maxvalu=m;
                }
            } System.out.println(maxvalu);
         
          t--;
      }
        
    }
       
    }
