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.*;
import java.util.*;
 

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);
        Unstoppable solver = new Unstoppable();
		int t=in.nextInt();
		while(t-->0)
        solver.solve(in, out);
        out.close();
    }
 
    static class Unstoppable {
        public void solve(InputReader in, PrintWriter out) {
         int n=in.nextInt();
         int a[]=new int[n];
         for(int i=0;i<n;i++){
             a[i]=in.nextInt();
         }
         int max=1;
         int b[][]=new int[n][n];
         int count[]=new int[27];
         for(int i=0;i<n;i++){
             Arrays.fill(count,0); max=1; count[a[i]]++; b[i][i]=1;
            for(int j=i+1;j<n;j++){
                count[a[j]]++; if(max<count[a[j]]) max=count[a[j]]; b[i][j]=max;
            }
         }max=1;
         int c=0;
         HashSet<Integer> h=new HashSet<>();
         for(int i=0;i<n;i++){
             int j=n,k=-1; if(h.contains(a[i])) continue;  h.add(a[i]); c=0;
             while(k<j){ int m=0,l=0;
                int flag=0;if(k+1<=j-1) flag=b[k+1][j-1]; if(max<c+flag) max=c+flag;  // out.println(k+" "+j+" "+max+" "+c+" "+b[k+1][j-1]);
                for(m=k+1;m<n;m++) { if(a[i]==a[m]) { c++; k=m; break; } } 
                for(l=j-1;l>=0;l--){ if(a[l]==a[i]) { c++; j=l; break; } }  
             }
         }
        // out.println(Arrays.toString(b[0]));
         out.println(max);
        }
 
    }

 
    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 long nextLong(){
            return Long.parseLong(next());
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
    }
}
 