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.*;

public class ThreeBlocksPalindrome {
    
    public static int getThreeBlockPalindrome(int b, char item, List<Character> str) {
        int best = 0;
        int pairs = 0;

        int left = 0;
        int right = str.size() - 1;
        while (left < right) {
            char l = str.get(left);
            char r = str.get(right);
            if (l == item && r == item) {
                pairs += 2;
                best = Math.max(best, pairs + b);
                left++;
                right--;
            }
            if (l != item) {
                b--;
                left++;
            }
            if (r != item) {
                b--;
                right--;
            }
        }

        return best;
    }

    static int solve(int[] nums) {
        int maxCount = 0;
        List<Integer> unique = new ArrayList<>();
        HashMap<Integer, Integer> counts = new HashMap<>();
        for (int num: nums) {
            if (!counts.containsKey(num)) unique.add(num);
            counts.merge(num, 1, Integer::sum);
            maxCount = Math.max(maxCount, counts.get(num));
        }

        for (int i = 0; i < unique.size(); i++) {
            for (int j = i+1; j < unique.size(); j++) {
                int num1 = unique.get(i);
                int num2 = unique.get(j);
                List<Character> str = new ArrayList<>();
                for (int num: nums) {
                    if (num == num1) str.add('1');
                    else if (num == num2) str.add('0');
                }
                int currentMax = Math.max(
                        getThreeBlockPalindrome(counts.get(num2), '1', str),
                        getThreeBlockPalindrome(counts.get(num1), '0', str)
                );
                maxCount = Math.max(maxCount, currentMax);
            }
        }

        return maxCount;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tests = sc.nextInt();
        for (int t = 0; t < tests; t++) {
            int n = sc.nextInt();
            int[] nums = new int[n];
            for (int i=0; i < n; i++) nums[i] = sc.nextInt();
            System.out.println(solve(nums));
        }
    }
}