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.*;

public class _1335_E1 {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int t = Integer.parseInt(in.readLine());
        for(int i = 0; i < t; i++) {
            int n = Integer.parseInt(in.readLine());
            int[] a = new int[n];
            StringTokenizer line = new StringTokenizer(in.readLine());
            int[][] prefix = new int[n][27];
            for(int j = 0; j < n; j++) {
                a[j] = Integer.parseInt(line.nextToken());
                for(int k = 0; k < 27; k++) {
                    prefix[j][k] = j > 0 ? prefix[j - 1][k] : 0;
                }
                prefix[j][a[j]]++;
            }
            int[][] suffix = new int[n][27];
            for(int j = n - 1; j >= 0; j--) {
                for(int k = 0; k < 27; k++) {
                    suffix[j][k] = j == n - 1 ? 0 : suffix[j + 1][k];
                }
                suffix[j][a[j]]++;
            }
            int max = 0;
            for(int l = 0; l < n; l++) {
                int[] count = new int[27];
                for(int r = l; r < n; r++) {
                    count[a[r]]++;
                    int maxmid = 0;
                    int maxsides = 0;
                    for(int j = 0; j < 27; j++) {
                        maxmid = Math.max(maxmid, count[j]);
                        maxsides = Math.max(maxsides, Math.min(l > 0 ? prefix[l - 1][j] : 0, r == n - 1 ? 0 : suffix[r + 1][j]));
                    }
                    max = Math.max(max, maxmid + (maxsides * 2));
                }
            }
            out.println(max);
        }
        in.close();
        out.close();
    }
}
