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.*;
import java.util.*;

public class E {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		StringBuilder answer = new StringBuilder();
		for (int tc = 1; tc <= t; ++tc) {
			int n = Integer.parseInt(br.readLine());
			StringTokenizer st = new StringTokenizer(br.readLine());
			int[] arr = new int[n];
			ArrayList<ArrayList<Integer>> index = new ArrayList<ArrayList<Integer>>();
			for (int i = 0; i <= 200; ++i) {
				index.add(new ArrayList<Integer>());
			}
			for (int i = 0; i < n; ++i) {
				arr[i] = Integer.parseInt(st.nextToken());
				index.get(arr[i]).add(i);
			}
			int max = 0;
			for (int i = 1; i <= 200; ++i) {
				int size = index.get(i).size() / 2;
				int fr = -1;
				int to = n;
				for (int k = 1; k <= 200; ++k) {
					int len = 0;
					if (k != i) {
						for (int idx : index.get(k)) {
							len++;
						}
					}
					if (len > max) {
						max = len;
					}
				}
				for (int j = 0; j < size; ++j) {
					fr = index.get(i).get(j);
					to = index.get(i).get(index.get(i).size() - j - 1);
					for (int k = 1; k <= 200; ++k) {
						int len = (j + 1) * 2;
						if (k != i) {
							for (int idx : index.get(k)) {
								if (fr < idx && idx < to) {
									len++;
								}
							}
						}
						if(len > max) {
							max = len;
						}
					}
				}
			}
			answer.append(max).append('\n');
		}
		System.out.println(answer);
	}
}