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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	public static int N;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		
		int T = Integer.parseInt(br.readLine());
		
		for(int testcase = 0; testcase < T; testcase++) {
			N = Integer.parseInt(br.readLine());
			st = new StringTokenizer(br.readLine());
			long[] arr = new long[N + 2];
			
			for(int i = 1; i <= N; i++) {
				arr[i] = Long.parseLong(st.nextToken());
			}
			arr[N + 1] = arr[N] - 1;
			
			System.out.println(solve(arr));
		}

	}
	
	public static long solve(long[] arr) {
		long max = 0;
		
		for(int i = 1; i <= N; i++) {
			if(i == 1) {
				if(arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) {
					max = Math.max(max, arr[i] * arr[i + 1]);
				}
			}
			else if(i == N) {
				if(arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) {
					max = Math.max(max, arr[i - 1] * arr[i]);
				}
			}
			else {
				if(arr[i - 1] <= arr[i] && arr[i] >= arr[i + 1]) {
					max = Math.max(max, arr[i - 1] * arr[i]);
					max = Math.max(max, arr[i] * arr[i + 1]);
				}
			}
		}
		
		return max;
	}

}
