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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException, InterruptedException {
        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nextInt() - 1;
            int[] cnt = new int[26];
            int ans = 0;
            for (int x : a) {
                cnt[x]++;
                ans = Math.max(ans, cnt[x]);
            }
            int[] left = new int[26], mid = new int[26];
            for (int l = 0; l < n; l++) {
                if (l > 0) left[a[l - 1]]++;
                Arrays.fill(mid, 0);
                int y = 0;
                for (int r = l; r < n; r++) {
                    mid[a[r]]++;
                    y = Math.max(y, mid[a[r]]);
                    for (int i = 0; i < 26; i++)
                        ans = Math.max(ans, y + 2*Math.min(left[i], cnt[i] - mid[i]-left[i]));
                }
            }
            out.println(ans);
        }
        out.flush();
        out.close();
    }

    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        boolean ready() throws IOException {
            return br.ready();
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        Scanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

        int nextChar() throws IOException {
            return next().charAt(0);
        }

        int[] nextIntArray(int n) throws IOException {
            int[] ans = new int[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

    }
}
