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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.StringTokenizer;

/**
 * http://codeforces.com/contest/1334/problem/D
 */
public class TaskE {

    public static void main(String[] arg) {
        final FastScanner in = new FastScanner(System.in);
        final PrintWriter out = new PrintWriter(System.out);
        final int T = in.nextInt();
        final StringBuilder sb = new StringBuilder();
        for(int t=0;t<T;t++){
            final int n = in.nextInt();
            final int [] a =in.readIntArr(n);
            sb.append(solution(n, a)).append("\n");
        }
        out.print(sb.toString());
        out.close();
        in.close();
    }

    private static int solution(int n, int[] a) {
        final int MAX = 27;
        int [] m = new int[MAX];
        int maxLen  = 0;
        for (int i = 0; i < n; i++) {
            m[a[i]]++;
            maxLen = Math.max(maxLen, m[a[i]]);
        }
        for(int i=1; i<MAX;i++){
            int j = m[i]/2;
            //search index
            for(int x=1;x<=j;x++) {
                int from = 0;
                int cnt = 0;
                while (from < n && cnt < x) {
                    if (a[from] == i) {
                        cnt++;
                    }
                    from++;
                }
                cnt = 0;
                int to = n - 1;
                while (to > 0 && cnt < x) {
                    if (a[to] == i) {
                        cnt++;
                    }
                    to--;
                }

                int y = findMax(a, from, to);
                maxLen = Math.max(x + y + x, maxLen);
            }

        }
        return maxLen;
    }

    private static int findMax(int [] a, int from, int to){
        int max  = 0;
        Map<Integer, Integer> m = new HashMap<>();
        for(int i=from;i<=to;i++){
            Integer count = m.getOrDefault(a[i],0);
            count++;
            m.put(a[i], count);
            max = Math.max(max, count);
        }
        return max;
    }


    private static int [] naive(final int n, final long l, final long r){
        final int [] result = new int[(int)(r-l+1)];
        long ind = 1;
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                if(ind>=l&&ind<=r){
                    result[(int)(ind-l)]=i;
                }
                ind++;
                if(ind>=l&&ind<=r){
                    result[(int)(ind-l)]=j;
                }
                ind++;
            }
        }
        if(ind>=l&&ind<=r){
            result[(int)(ind-l)]=1;
        }
        ind++;
        return result;
    }

    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

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

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

        int[] readIntArr(int n) {
            int[] result = new int[n];
            for (int i = 0; i < n; i++) {
                result[i] = Integer.parseInt(next());
            }
            return result;
        }

        void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

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

    }
}
