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 Solution {
    static int arr[][];
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        int t=sc.nextInt();
        StringBuilder sb=new StringBuilder();
        while (t-->0){
            int n=sc.nextInt();
            int arr[]=new int[n+1];
            for (int i=1;i<=n;i++){
                arr[i]=sc.nextInt()-1;
            }
            int sum[][]=new int[n+1][26];
            for (int i=1;i<=n;i++){
                sum[i][arr[i]]++;
                for (int j=0;j<26;j++){
                    sum[i][j]+=sum[i-1][j];
                }
            }
            int ams=1;
            for (int i = 1; i <= n; i++) {
                for (int j = i + 1; j <= n; j++) {
                    int max=0;
                    for (int a=0;a<26;a++){
                        max=Math.max(max,2*Math.min(sum[i][a],sum[n][a]-sum[j-1][a]));
                    }
                    int m=0;
                    for (int b=0;b<26;b++){
                        m=Math.max(m,sum[j-1][b]-sum[i][b]);
                    }
                    ams=Math.max(ams,max+m);
                }
            }
//            System.out.println(ams);
            sb.append(ams+"\n");
        }
        System.out.print(sb);
    }
    static void c(int i,int j){
        if (arr[i][j]==1)arr[i][j]++;
        else arr[i][j]--;
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}