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.Scanner;

/**
 * Codeforces Problem 1335E2
 */
public class Solution {

	public static void main(String args[]) {
		try (Scanner scanner = new Scanner(System.in);) {
			int t = scanner.nextInt();
			for (int i = 0; i < t; i++) {
				String solution = solveNext(scanner);
				System.out.println(solution);
			}
		}
		System.exit(0);
	}

	public static String solveNext(Scanner scanner) {
		int numItems = scanner.nextInt();
		int[] items = new int[numItems];
		for (int i = 0; i < numItems; i++) {
			items[i] = scanner.nextInt();
		}
		return new Solution(numItems, items).solve();
	}

	static final int A_MAX = 200;

	final int numItems;
	final int[] items;

	public Solution(int numItems, int[] items) {
		this.numItems = numItems;
		this.items = items;
	}

	private static class Node {

		int[] leftCounts;
		Node prev;
		Node next;

	}

	private int getOptimum(int a, Node head, Node tail, int size) {
		int iterCount = size / 2;
		int max = 0;
		for (int i = 1; i <= iterCount; i++) {
			int maxMiddle = 0;
			for (int j = 0; j < A_MAX; j++) {
				if (j == a) {
					continue;
				}
				maxMiddle = Math.max(maxMiddle, tail.leftCounts[j] - head.leftCounts[j]);
			}
			max = Math.max(max, maxMiddle + (2 * i));
			head = head.next;
			tail = tail.prev;
		}
		return max;
	}

	public String solve() {
		Node[] headNodes = new Node[A_MAX];
		Node[] tailNodes = new Node[A_MAX];
		int[] prevCount = new int[A_MAX];
		int max = 0;
		for (int i = 0; i < this.numItems; i++) {
			int a = this.items[i] - 1;
			Node currNode = new Node();
			currNode.leftCounts = prevCount;

			if (headNodes[a] == null) {
				headNodes[a] = currNode;
				tailNodes[a] = currNode;
			} else {
				tailNodes[a].next = currNode;
				currNode.prev = tailNodes[a];
				tailNodes[a] = currNode;
			}

			prevCount = new int[A_MAX];
			System.arraycopy(currNode.leftCounts, 0, prevCount, 0, A_MAX);
			prevCount[a]++;
			max = Math.max(max, prevCount[a]);
		}

		int[] totalCounts = prevCount;
		for (int i = 0; i < A_MAX; i++) {
			Node headNode = headNodes[i];
			if (headNode == null) {
				continue;
			}
			Node tailNode = tailNodes[i];
			max = Math.max(max, this.getOptimum(i, headNode, tailNode, totalCounts[i]));
		}

		return Integer.toString(max);
	}

}
