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.io.*;
import java.util.*;
 
public class Cherry {

    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        for (int i = 1; i <= t; i++)  {
        	int n = in.nextInt();
        	long[] val = new long[n];
        	for (int k = 0; k < n; k++)  {
        		val[k] = in.nextLong();
        	}
        	
        	if (n == 2)  {
        		out.println(val[0] * val[1]);
        	}
        	else  {
            	long maxproduct = 0;
            	for (int k = 1; k < n - 1; k++)  {
            		if ((val[k] >= val[k - 1]) && (val[k] >= val[k + 1]))  {
            			long temp1 = val[k] * val[k - 1];
            			long temp2 = val[k] * val[k + 1];
            			
            			if (Math.max(temp1, temp2) > maxproduct)  {
            				maxproduct = Math.max(temp1, temp2);
            			}
            		}
            	}
            	
    			maxproduct = Math.max(maxproduct, val[0] * val[1]);
    			maxproduct = Math.max(maxproduct, val[n - 2] * val[n - 1]);
            	
            	out.println(maxproduct);
        	}
        }
        out.close();
    }
    
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try { 
                    st = new StringTokenizer(br.readLine());                               
                } catch (IOException e) {
                    // noop
                }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble()  {
            return Double.parseDouble(next());
        }
    }

}