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.lang.*;
import java.io.*;
 
public class e1
{
    static int getf(int[][] f, int c, int s, int e){
        if(e<s) return 0;
        int fr = f[c][e];
        if(s>0){
            fr -= f[c][s-1];
        }
        return fr;
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		MyScanner in = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
		int T=in.nextInt();
		int nc = 26;
		for(int t=0;t<T;t++){
			int n = in.nextInt();
			int[][] f = new int[nc+1][n];
			int[] a = new int[n];
			for(int i=0;i<n;i++){
			    a[i] = in.nextInt();
			    for(int c=1;c<=nc;c++){
			        int p = 0;
			        if(i>0){
			            p = f[c][i-1];
			        }
			        if(a[i]==c){
			            p++;
			        }
			        f[c][i] = p;
			    }
			}
			int ans = 0;
			for(int i=0;i<n;i++){
			    for(int j=i;j<n;j++){
			        int maxif=0;
			        int maxof=0;
			        for(int c=1;c<=nc;c++){
			            int cif = getf(f, c, i, j);
			            if(cif>maxif) maxif=cif;
			            int lf = getf(f, c, 0, i-1);
			            int rf = getf(f, c, j+1, n-1);
			            int cof = Math.min(lf, rf);
			            if(cof>maxof) maxof = cof;
			        }
			        //out.println("i,j,maxif,maxof:"+i+","+j+","+maxif+","+maxof);
			        int cl = maxif + maxof*2;
			        if(cl>ans) ans = cl;
			    }
			}
			out.println(ans);
		}
		out.close();
	}
	public static PrintWriter out;
      
   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         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;
      }
 
   }
}