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.InputStreamReader;
import java.util.StringTokenizer;

public class ProblemE {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		FastReader s = new FastReader();
		int t = s.nextInt();
		for(int a = 0;a<t;a++) {
			int n = s.nextInt();
			
			int[] arr = new int[n];
			
			for(int i=0;i<n;i++)
				arr[i] = s.nextInt();
			
			int[][] cum = new int[n][201];
			
			for(int i=0;i<n;i++)
				for(int j=1;j<=200;j++) {
					
					if(i==0)
					{
						if(j==arr[0])
							cum[i][j] = 1;
					}
					else if(j == arr[i])
						cum[i][j] = cum[i-1][j] + 1;
					
					else
						cum[i][j] = cum[i-1][j];
				}
			
			
			int max = 1;
			
			for(int i=1;i<=200;i++) {
				max = Math.max(max, findMax(cum, arr, i));
			}
			
			System.out.println(max);
		}

	}
	
	public static int findMax(int[][] cum, int[] arr, int val) {
		
		
		int max = 0;
		int n = arr.length;
		int i=0, j = n-1;
		//System.out.println("hey"+val);
		int count = 0;
		while(true) {
			while(i < n && arr[i] != val)
				i++;
			//System.out.println(i);
			while(j > 0 && arr[j] != val)
				j--;
			
			if(i == n || j < 0 || i >= j)
				break;
			
			count++;
			int mx = 0;
			
			for(int k=1;k<=200;k++) {
				int df = cum[j-1][k] - cum[i][k];
				
				mx = Math.max(mx, df);
			}
			i++;
			j--;
			
			max = Math.max(max, mx + 2*count);
		}
		//System.out.println("returning");
		return max;
	}
	
	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;
        }
    }

}
