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

public class B {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();

		while (t-- > 0) {
			int n = sc.nextInt();
			int cnt[][] = new int[26][n];
			int arr[] = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = sc.nextInt() - 1;

				for (int j = 0; j < 26; j++) {
					if (j == arr[i]) {
						cnt[j][i] = (i == 0 ? 0 : cnt[j][i - 1]) + 1;
					} else {
						cnt[j][i] = (i == 0 ? 0 : cnt[j][i - 1]);
					}
				}
			}
			// 1 1
			// [1,2]
			int left[][] = new int[n][n];
			for (int j = 0; j < n; j++) {
				for (int k = j; k < n; k++) {
					int maxLeft = 0;
					for (int l = 0; l < 26; l++) {
						int ll = 2 * Math.min(j == 0 ? 0 : cnt[l][j - 1], cnt[l][n - 1] - cnt[l][k]);
						maxLeft = Math.max(maxLeft, ll);
					}
					left[j][k] = maxLeft;

				}
			}
			int max = 0;
			for (int i = 0; i < 26; i++) {
				for (int j = 0; j < n; j++) {
					for (int k = j; k < n; k++) {
						int middle = cnt[i][k] - (j == 0 ? 0 : cnt[i][j - 1]);

						max = Math.max(max, middle + left[j][k]);

					}
				}
			}
			System.out.println(max);
		}
	}
}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s) {
		br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

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

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

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public double nextDouble() throws IOException {
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if (x.charAt(0) == '-') {
			neg = true;
			start++;
		}
		for (int i = start; i < x.length(); i++)
			if (x.charAt(i) == '.') {
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			} else {
				sb.append(x.charAt(i));
				if (dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg ? -1 : 1);
	}

	public boolean ready() throws IOException {
		return br.ready();
	}

}