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 ProblemE1 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int cases = s.nextInt();
        for (int x = 0; x < cases; x++) {
            int n = s.nextInt();
            int[] ar = new int[n];
            for (int i = 0; i < n; i++) {
                ar[i] = s.nextInt();
            }
            int[][] pre = new int[26][n];
            for (int i = 0; i < 26; i++) {
                if (ar[0] == i+1) {
                    pre[i][0] = 1;
                }
                else {
                    pre[i][0] = 0;
                }
                for (int j = 1; j < n; j++) {
                    if (ar[j] == i+1) {
                        pre[i][j] = pre[i][j-1] + 1;
                    }
                    else {
                        pre[i][j] = pre[i][j-1];
                    }
                }
            }
            int max = 1;
            for (int i = 0; i < 26; i++) {
                int p1 = 0;
                int p2 = n-1;
                int counter = 0;
                while (p1 < p2) {
                    if (ar[p1] == i+1) {
                        counter++;
                        while (ar[p2] != i+1 && p1 < p2) {
                            p2--;
                        }
                        if (p1 < p2) {
                            int curMax = 0;
                            for (int j = 0; j < 26; j++) {
                                int c = pre[j][p2] - pre[j][p1] + counter * 2;
                                if (j == i) {c--;}
                                if (c > curMax) {
                                    curMax = c;
                                }
                            }
                            if (curMax > max) {
                                max = curMax;
                            }
                        }
                        p2--;
                    }
                    p1++;
                }
            }
            System.out.println(max);
        }
    }
}
