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.*;
import java.lang.*;
import static java.lang.Math.*;

// Sachin_2961 submission //
public class Codeforces {

    public void solve() {
        int n = fs.nInt();
        int[]ar = new int[n];
        for(int i=0;i<n;i++){
            ar[i] = fs.nInt();
        }
        long max = 0,ans=0,min = Long.MAX_VALUE;
        int ind = -1;
        for(int i=0;i<n;i++){
            if(ar[i] > max){
                max = ar[i];
                ind = i;
            }
        }
        int l = ind-1;
        while (l >= 0){
            min = Math.min(min,ar[l]);
            ans = Math.max(max*min,ans);
            l--;
        }
        l = ind+1;
        min = max;
        while (l < n ){
            min = Math.min(min,ar[l]);
            ans = Math.max(max*min,ans);
            l++;
        }
        out.println(ans);
    }
    static boolean multipleTestCase = true; static FastScanner fs; static PrintWriter out;
    public void run(){
        fs = new FastScanner();
        out = new PrintWriter(System.out);
        int tc = (multipleTestCase)?fs.nInt():1;
        while (tc-->0)solve();
        out.flush();
        out.close();
    }
    public static void main(String[]args){
        try{
            new Codeforces().run();
        }catch (Exception e){
            e.printStackTrace();
        }

    }
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String n() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        String Line()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
        int nInt() {return Integer.parseInt(n()); }
        long nLong() {return Long.parseLong(n());}
        double nDouble(){return Double.parseDouble(n());}
        int[]aI(int n){
            int[]ar = new int[n];
            for(int i=0;i<n;i++)
                ar[i] = nInt();
            return ar;
        }
    }
    public static void sort(int[] arr){
        ArrayList<Integer> ls = new ArrayList<Integer>();
        for(int x: arr)
            ls.add(x);
        Collections.sort(ls);
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
}