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.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;

public class Cherry {

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	public static void main(String[] args) {

		FastScanner f = new FastScanner();
		PrintWriter p = new PrintWriter(System.out);
		int t = f.nextInt();
		while (t-- > 0) {
			int n = f.nextInt();
			long arr[] = new long[n];
			long max = Long.MIN_VALUE;
			for (int i = 0; i < n; i++) {
				arr[i] = f.nextLong();
			}
			if (n == 2) {
				max = arr[0] * arr[1];
			}
			for (int i = 1; i < n - 1; i++) {
				long l = arr[i] * arr[i - 1];
				long r = arr[i] * arr[i + 1];
				max = Math.max(max, Math.max(l, r));
			}

			p.println(max);

		}
		p.close();

	}

	// will work till array size of 65537---fastest sorting time
//	static long[] radixSort(long[] f) {
//		return radixSort(f, f.length);
//	}
//
//	static long[] radixSort(long[] f, int n) {
//		long[] to = new long[n];
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] >>> 16 & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] >>> 16 & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] >>> 32 & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] >>> 32 & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		{
//			int[] b = new int[65537];
//			for (int i = 0; i < n; i++)
//				b[1 + (int) (f[i] >>> 48 & 0xffff)]++;
//			for (int i = 1; i <= 65536; i++)
//				b[i] += b[i - 1];
//			for (int i = 0; i < n; i++)
//				to[b[(int) (f[i] >>> 48 & 0xffff)]++] = f[i];
//			long[] d = f;
//			f = to;
//			to = d;
//		}
//		return f;
//	}

	// Function to sort an array using quick sort algorithm.
	static void quickSort(long arr[], int low, int high) {
		if (low < high) {
			int p = partition(arr, low, high);
			quickSort(arr, low, p - 1);
			quickSort(arr, p + 1, high);
		}
	}

	static int partition(long arr[], int low, int high) {
		long pivot = arr[high];
		int i = low - 1;
		for (int j = low; j < high; j++) {
			if (arr[j] < pivot) {
				i++;
				long temp = arr[i];
				arr[i] = arr[j];
				arr[j] = temp;
			}
		}
		long temp = arr[i + 1];
		arr[i + 1] = arr[high];
		arr[high] = temp;

		return i + 1;
	}

	/// Ceils
	static long ceil(long x, long m) {
		long res = x / m;
		if (x % m != 0) {
			res++;
		}
		return res;
	}

	// ------------------------------------------------------------------------------------------------
	// makes the prefix sum array
	static long[] prefixSum(long arr[], int n) {
		long psum[] = new long[n];
		psum[0] = arr[0];
		for (int i = 1; i < n; i++) {
			psum[i] = psum[i - 1] + arr[i];
		}
		return psum;
	}
	// ------------------------------------------------------------------------------------------------

	// makes the suffix sum array
	static long[] suffixSum(long arr[], int n) {
		long ssum[] = new long[n];
		ssum[n - 1] = arr[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			ssum[i] = ssum[i + 1] + arr[i];
		}
		return ssum;
	}

//------------------------------------------------------------------------------------------
	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
	// OVERHEADS
	static long m = (long) (1e9 + 7);

	static long binPower(long a, long n, long m) {
		if (n == 0)
			return 1;
		long res = 1;
		while (n > 0) {
			if ((n & 1) != 0) {
				res *= a;
			}
			a *= a;
			n >>= 1;
		}
		return res;
	}

//-------------------------------------------------------------------------------------------

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

//------------------------------------------------------------------------------------------

	// lcm
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

//------------------------------------------------------------------------------------------
	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
	// O(LOGn)
	static int setBits(int n) {
		int count = 0;
		while (n > 0) {
			n = n & (n - 1);
			count++;
		}
		return count;
	}
//------------------------------------------------------------------------------------------

//------------------------------------------------------------------------------------------
	// 0 based indexing
	static boolean KthBitSet(int n, int k) {
		int mask = 1;
		mask = mask <<= k;
		if ((mask & n) != 0)
			return true;
		else
			return false;
	}

//------------------------------------------------------------------------------------------

	// EXTENDED EUCLIDEAN THEOREM
	// TO REPRESENT GCD IN TERMS OF A AND B
	// gcd(a,b) = a.x + b.y where x and y are integers

	static long x = -1;
	static long y = -1;

	static long gcdxy(long a, long b) {
		if (b == 0) {
			x = 1;
			y = 0;
			return a;
		} else {
			long d = gcdxy(b, a % b);
			long x1 = y;
			long y1 = x - (a / b) * y;
			x = x1;
			y = y1;

			return d;
		}
	}

//-------------------------------------------------------------------------------------------------F
}
