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 implements Runnable{
    FastScanner sc;
    PrintWriter pw;
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args)
    {
        new Thread(null,new Solution(),"codeforces",1<<25).start();
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
       solve();
        pw.flush();
        pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    public long ppow(long a,long b,long mod)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1)
            tmp*=a;
            a*=a;
            a%=mod;
            tmp%=mod;
            b>>=1;
        }
        return (tmp*a)%mod;
    }
    public  int gcd(int x,int y)
    {
        return y==0?x:gcd(y,x%y);
    }
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////
   public class Pair implements Comparable<Pair>{
       int f;
       int n;
       Pair(int a,int b)
       {
           n=a;
           f=b;
       }
       public int compareTo(Pair j)
       {
           return f-j.f;
       }
   }
    void solve()
   {
       int t=sc.ni();
       while(t-->0)
       {
          int n=sc.ni();
          int[] arr=new int[n];
          for(int i=0;i<n;i++)
          arr[i]=sc.ni();
          int[][] prr=new int[n][27];
          int[][] srr=new int[n][27];
          srr[n-1][arr[n-1]]++;
          prr[0][arr[0]]++;
          for(int i=1;i<n;i++)
          {
              for(int j=1;j<=26;j++)
              prr[i][j]=prr[i-1][j];
            prr[i][arr[i]]++;
          }
          for(int i=n-2;i>=0;i--)
          {
              for(int j=1;j<=26;j++)
                srr[i][j]=srr[i+1][j];
                srr[i][arr[i]]++;
          }
          int ans=0;
          for(int j=1;j<=26;j++)
          ans=Math.max(ans,prr[n-1][j]);
          for(int i=1;i<n-1;i++)
          {
             for(int j=i;j<n-1;j++)
             {
                 int[] crr=new int[27];
                 int m1=0,m2=0;
                 for(int k=1;k<=26;k++)
                 {
                    crr[k]=prr[j][k]-prr[i-1][k];
                    m1=Math.max(m1,crr[k]);
                    m2=Math.max(Math.min(prr[i-1][k],srr[j+1][k]),m2);
                 }
                 ans=Math.max(ans,m1+m2+m2);
             }
          }
          pw.println(ans);
       }
   }
}