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
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int t = in.nextInt();
            while (t > 0) {
                t--;
                int n = in.nextInt();
                int[] a = new int[n];
                int max = 26;
                int[] c = new int[max];
                for (int i = 0; i < n; i++) {
                    a[i] = in.nextInt();
                    c[a[i] - 1]++;
                }
                int ans = 0;
                for (int v : c)
                    ans = Math.max(ans, v);
                int[][] pref = new int[max][n];
                for (int i = 0; i < max; i++)
                    pref[i] = pref(a, n, i + 1);
                int[][] z = new int[n][n];
                for (int j = 0; j < n; j++) {
                    z[j] = new int[n];
                    for (int k = 0; k < n; k++) {
                        int mx = 0;
                        for (int i = 1; i <= max; i++) {
                            mx = Math.max(mx, sum(pref, i, j, k));
                        }
                        z[j][k] = mx;
                    }
                }
                for (int v = 1; v <= max; v++) {
                    for (int i = 0; i + 2 < n; i++) {
                        int l = sum(pref, v, 0, i);
                        int min = l + 1;
                        int rest = n - 1 - (i + 1) + 1;
                        if (rest < min)
                            break;
                        else {
                            for (int j = n - 1; j > i + 1; j--) {
                                int r = sum(pref, v, j, n - 1);
                                if (l == r) {
                                    int mx = z[i + 1][j - 1];
                                    ans = Math.max(ans, l + mx + r);
                                } else {
                                    if (r > l)
                                        break;
                                }
                            }
                        }
                    }
                }
                out.println(ans);
            }
        }

        int sum(int[][] pref, int v, int l, int r) {
            if (l > r)
                return 0;
            v--;
            if (l == 0)
                return pref[v][r];
            else
                return pref[v][r] - pref[v][l - 1];
        }

        int[] pref(int[] ar, int n, int v) {
            int[] pref = new int[n];
            if (ar[0] == v)
                pref[0] = 1;
            for (int i = 1; i < n; i++)
                if (ar[i] == v)
                    pref[i] = pref[i - 1] + 1;
                else
                    pref[i] = pref[i - 1];
            return pref;
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}

