The only difference between easy and hard versions is constraints.

You are given a sequence a consisting of n positive integers.

Let's define a three blocks palindrome as the sequence, consisting of at most two distinct elements (let these elements are a and b, a can be equal b) and is as follows: [\underbrace{a, a, ..., a}_{x}, \underbrace{b, b, ..., b}_{y}, \underbrace{a, a, ..., a}_{x}]. There x, y are integers greater than or equal to 0. For example, sequences [], [2], [1, 1], [1, 2, 1], [1, 2, 2, 1] and [1, 1, 2, 1, 1] are three block palindromes but [1, 2, 3, 2, 1], [1, 2, 1, 2, 1] and [1, 2] are not.

Your task is to choose the maximum by length subsequence of a that is a three blocks palindrome.

You have to answer t independent test cases.

Recall that the sequence t is a a subsequence of the sequence s if t can be derived from s by removing zero or more elements without changing the order of the remaining elements. For example, if s=[1, 2, 1, 3, 1, 2, 1], then possible subsequences are: [1, 1, 1, 1], [3] and [1, 2, 1, 3, 1, 2, 1], but not [3, 2, 3] and [1, 1, 1, 1, 2].

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2000) — the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 26), where a_i is the i-th element of a. Note that the maximum value of a_i can be up to 26.

It is guaranteed that the sum of n over all test cases does not exceed 2000 (∑ n ≤ 2000).

Output

For each test case, print the answer — the maximum possible length of some subsequence of a that is a three blocks palindrome.

Example

Input


6
8
1 1 2 2 3 2 1 1
3
1 3 3
4
1 10 10 1
1
26
2
2 1
3
1 1 1


Output


7
2
4
1
1
3

import java.io.IOException;
import java.util.Scanner;

public final class Main
{
	public static void main(String args[]) throws IOException
	{
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		in.nextLine();
		while (t-- > 0)
		{
			int n = in.nextInt();
			int[][] cntArr = new int[26][n];
			int max = 0;
			for (int i = 0; i < n; i++)
			{
				int v = in.nextInt() - 1;
				for (int j = 0; j < 26 && i > 0; j++)
				{
					cntArr[j][i] = cntArr[j][i - 1];
				}
				cntArr[v][i]++;
				max = Math.max(max, cntArr[v][i]);
			}

			for (int l = 1; l < n - 1; l++)
			{
				for (int r = l; r < n - 1; r++)
				{
					int cntin = 0;
					for (int i = 0; i < 26; i++)
					{
						cntin = Math.max(cntin, cntArr[i][r] - cntArr[i][l - 1]);
					}
					int cntout = 0;
					for (int i = 0; i < 26; i++)
					{
						int tmp = Math.min(cntArr[i][l - 1], cntArr[i][n - 1] - cntArr[i][r]);
						cntout = Math.max(cntout, tmp);
					}
					max = Math.max(max, 2 * cntout + cntin);
				}
			}
			System.out.println(max);
		}
	}
}
