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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Washoum
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        inputClass in = new inputClass(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        E1ThreeBlocksPalindromeEasyVersion solver = new E1ThreeBlocksPalindromeEasyVersion();
        solver.solve(1, in, out);
        out.close();
    }

    static class E1ThreeBlocksPalindromeEasyVersion {
        public void solve(int testNumber, inputClass sc, PrintWriter out) {
            int t = sc.nextInt();
            while (t > 0) {
                t--;
                int n = sc.nextInt();
                int[][] a = new int[n][26];
                for (int i = 0; i < n; i++) {
                    int x = sc.nextInt() - 1;
                    if (i > 0) {
                        for (int j = 0; j < 26; j++) {
                            a[i][j] = a[i - 1][j];
                        }
                    }
                    a[i][x]++;
                }
                int best = 0;
                for (int i = 0; i < 26; i++) {
                    best = Math.max((a[n - 1][i] / 2) * 2, best);
                }
                for (int i = 0; i < n; i++) {
                    for (int j = i; j < n; j++) {
                        int[] mid = new int[26];
                        int idx1 = 0;
                        int idx2 = -1;
                        if (i == 2 && j == 5) {
                            out.println();
                        }
                        for (int k = 0; k < 26; k++) {
                            if (i > 0) {
                                mid[k] = a[j][k] - a[i - 1][k];
                            } else {
                                mid[k] = a[j][k];
                            }
                            if (mid[k] > mid[idx1]) {
                                idx1 = k;
                            } else {
                                if (idx2 == -1) {
                                    idx2 = k;
                                } else if (mid[k] > mid[idx2]) {
                                    idx2 = k;
                                }
                            }
                        }
                        int[] right = new int[26];
                        for (int k = 0; k < 26; k++) {
                            right[k] = a[n - 1][k] - a[j][k];
                        }
                        int[] left = new int[26];
                        for (int k = 0; k < 26; k++) {
                            if (i > 0) {
                                left[k] = a[i - 1][k];
                            }
                        }
                        int nb = 0;
                        for (int k = 0; k < 26; k++) {
                            nb = 2 * Math.min(left[k], right[k]);
                            if (k == idx1) {
                                nb += mid[idx2];
                            } else {
                                nb += mid[idx1];
                            }
                            best = Math.max(best, nb);
                        }
                    }
                }
                out.println(best);
            }
        }

    }

    static class inputClass {
        BufferedReader br;
        StringTokenizer st;

        public inputClass(InputStream in) {

            br = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

