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.*;
import java.io.*;
public class e {
	static final int max = 26;
	static int n;
	static int[] arr;
	static int[][] pre;
	static int[][] idxForNum;
	public static void main(String[] args) {
		FS sc = new FS();
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		for(int tt = 1; tt <= t; ++tt) {
			n = sc.nextInt();
			arr = new int[n];
			pre = new int[max][n];
			idxForNum = new int[max][n + 1];
			for(int[] z : idxForNum) Arrays.fill(z, -2);
			for(int a = 0; a < max; ++a) idxForNum[a][0] = -1;

			for(int i = 0; i < n; ++i) {
				int in = arr[i] = sc.nextInt() - 1;

				// update all pres
				if(i > 0) {
					for(int a = 0; a < max; ++a) pre[a][i] = pre[a][i - 1];
				}

				// update current pre
				++pre[in][i];

				idxForNum[in][pre[in][i]] = i;
			}
			
			int out = 0;
			for(int r = 0; r < n; ++r) {
				for(int a = 0; a < max; ++a) {
					int x = getRange(r + 1, n - 1, a);
					
					int l = idxForNum[a][x] + 1;
					
					if(l == -1 || l > r) continue;
					
					for(int b = 0; b < max; ++b) {
						int y = getRange(l, r, b);
						out = Math.max(out, x + y + x);
					}
				}
			}
			
			pw.println(out);
		}
		pw.flush();
	}
	static int getRange(int l, int r, int a) {
		if(l > r) return 0;
		int out = pre[a][r];
		if(l > 0) out -= pre[a][l - 1];
		return out;
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}