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

public class cp{
    static BufferedReader br;
    static StringTokenizer st;

    public static void main(String[] args)throws IOException{

        br = new BufferedReader(new InputStreamReader(System.in));
        List<Long> answer = new ArrayList<Long>();

        for(int t=Integer.parseInt(br.readLine()); t>0; t--){

            st = readLine();
            int n = tokenInt();

            st = readLine();
            int[] arr = readIntArray(n);

            answer.add(new Solver().solve(n, arr));

        }

        for (long ans : answer ) System.out.println(ans);

    }

    static StringTokenizer readLine() throws IOException { return new StringTokenizer(br.readLine()); }
    static int tokenInt() { return Integer.parseInt(st.nextToken()); }
    static long tokenLong(){ return Long.parseLong(st.nextToken()); }
    static double tokenDouble(){ return Double.parseDouble(st.nextToken()); }
    static char[] tokenChar(){ return st.nextToken().toCharArray(); }
    static int[] readIntArray(int n) {
        int[] a=new int[n];
        for (int i=0; i<n; i++) a[i]=tokenInt();
        return a;
    }

    static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b);}
    static int lcm(int a, int b) { return (a * b) / gcd(a,b);}

    static void printArr(int[] arr){
        for (int i=0; i<arr.length; i++) System.out.print(arr[i] + " ");
        System.out.println();
    }

    static void printArr(char[] arr){
        for (int i=0; i<arr.length; i++) System.out.print(arr[i] + " ");
        System.out.println();
    }

    static void printArr2d(int[][] arr){
        for (int[] ar : arr) printArr(ar);
    }

    static void printArr2d(char[][] arr){
        for (char[] ar : arr) printArr(ar);
    }

}


class Solver{
    public long solve(int n, int[] arr) throws IOException{

        long prod = 0;
        if (n > 2){
            for (int i=1; i<n; i++) {
                prod = max((long) arr[i-1] * arr[i], prod);
            }
        } else{
            prod = arr[0] * arr[1];
        }

        return prod;
    }
}

/*
1
4
11 25 37 8
*/
