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
//package threeblockspalindromeeasyversion;

import java.util.*;
import java.io.*;

public class threeblockspalindromeeasyversion {
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(fin.readLine());
		StringBuilder fout = new StringBuilder();
		while(t-- > 0) {
			int n = Integer.parseInt(fin.readLine());
			int[] nums = new int[n];
			int[] numNums = new int[26];
			int[][] pfx = new int[26][n];
			StringTokenizer st = new StringTokenizer(fin.readLine());
			for(int i = 0; i < n; i++) {
				nums[i] = Integer.parseInt(st.nextToken()) - 1;
				numNums[nums[i]] ++;
				pfx[nums[i]][i] ++;
				if(i != 0) {
					for(int j = 0; j < 26; j++) {
						pfx[j][i] += pfx[j][i - 1];
					}
				}
			}
//			for(int[] i : pfx) {
//				for(int j : i) {
//					System.out.print(j + " ");
//				}
//				System.out.println();
//			}
//			for(int i : numNums) {
//				System.out.print(i + " ");
//			}
			//System.out.println();
			int ans = 0;
			for(int i = 0; i < 26; i++) {
				int cur = 0;
				int low = 0;
				int high = n - 1;
				for(int j = 1; j <= numNums[i] / 2; j++) {
					cur += 2;
					while(pfx[i][low] != j) {
						low ++;
					}
					if(j != 1) {
						high --;
					}
					while(nums[high] != i) {
						high --;
					}
					int max = 0;
					for(int k = 0; k < 26; k++) {
						max = Math.max(max, pfx[k][high - 1] - pfx[k][low]);
					}
					//System.out.println(max + " " + low + " " + high);
					ans = Math.max(max + cur, ans);
				}
			}
			for(int i = 0; i < 26; i++) {
				ans = Math.max(ans, numNums[i]);
			}
			fout.append(ans).append("\n");
		}
		System.out.print(fout);
	}
}
