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.util.Scanner;

public class E1 {

	public static void main(String[] args) {
		
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		for (int z = 0; z < t; z++) {
			int n = s.nextInt();
			int[] arr = new int[n];
			int[][] count = new int[27][n];
			int singleMax = 0;
			for (int i = 0; i < n; i++) {
				arr[i] = s.nextInt();
				if (i != 0) {
					for (int j = 0; j < 27; j++) {
						count[j][i] += count[j][i - 1];
					}
					count[arr[i]][i]++;
				} else {
					count[arr[i]][i]++;
				}
				singleMax = Math.max(singleMax, count[arr[i]][i]);
			}
			int max = singleMax;
			for (int a = 1; a < 27; a++) {
				for (int b = 1; b < 27; b++) {
					int end = n - 1;
					int endCount = 0;
					if (arr[end] == a) {
						endCount++;
					}
					int bCount = count[b][end];
					if (arr[end] == b) {
						bCount--;
					}
					int startCount = 0;
					for (int start = 0; start < n; start++) {
						if (arr[start] == a) {
							startCount++;
						}
						if (arr[start] == b) {
							bCount--;
						}
						while (end > 0 && endCount < startCount) {
							end--;
							if (arr[end] == a) {
								endCount++;
							}
							if (arr[end] == b) {
								bCount--;
							}
						}
						if (start < end) {
							max = Math.max(max, startCount * 2 + bCount);
						} else {
							break;
						}
					}
				}
			}
			System.out.println(max);
		}
		
	}
	
}
