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.*;
import java.text.*;

public class E1335 {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] arr = new int[n];
			int[] cnt = new int[201];
			for (int i = 0; i < arr.length; i++) {
				arr[i] = sc.nextInt();
				cnt[arr[i]]++;
			}
			int ans = 0;
			for (int a = 1; a <= 200; a++) {
				if (cnt[a] == 0)
					continue;
				ans = Math.max(ans, cnt[a]);
				if (cnt[a] == 1)
					continue;
				int l = 0;
				int r = n - 1;
				int[] tempcnt = cnt.clone();
				int[] cntcnt = new int[n + 1];
				for (int i = 0; i < cnt.length; i++) {
					cntcnt[cnt[i]]++;
				}
				int max = -1;
				for (int i = n; i >= 0; i--)
					if (cntcnt[i] != 0) {
						max = i;
						break;
					}
				int x = 0;
				loop: while (l < r) {
					while (arr[l] != a) {
						cntcnt[tempcnt[arr[l]]--]--;
						cntcnt[tempcnt[arr[l]]]++;
						l++;
						if (l == r)
							break loop;
					}
					while (arr[r] != a) {
						cntcnt[tempcnt[arr[r]]--]--;
						cntcnt[tempcnt[arr[r]]]++;
						r--;
						if (l == r)
							break loop;
					}
					cntcnt[tempcnt[arr[l]]--]--;
					cntcnt[tempcnt[arr[l]]]++;
					l++;
					if (l == r)
						break loop;
					cntcnt[tempcnt[arr[r]]--]--;
					cntcnt[tempcnt[arr[r]]]++;
					r--;
					
					while (cntcnt[max] == 0)
						max--;
					x += 2;
					ans = Math.max(ans, x + max);
				}
			}
			pw.println(ans);
		}
		pw.close();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		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 {
			return Double.parseDouble(next());
		}

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < arr.length; i++)
				arr[i] = nextInt();
			return arr;
		}

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

	}
}
