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 CJ3 {

	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		while (t>0)
		{
			int n=Integer.parseInt(br.readLine());
			int arr[]=new int[n];
			HashMap<Integer, ArrayList<Integer>> map=new HashMap<>();
			String inp[]=br.readLine().split(" ");
			for (int i=0;i<n;i++)
			{
				arr[i]=Integer.parseInt(inp[i]);
				if (map.containsKey(arr[i]))
					map.get(arr[i]).add(i);
				else
				{
					ArrayList<Integer> temp =  new ArrayList<>();
					temp.add(i);
					map.put(arr[i], temp);
				}
			}
			int maxlen=1;
			for (int i=1;i<=26;i++)
			{
				if (!map.containsKey(i))
					continue;
				ArrayList<Integer> indexesa=map.get(i);
				maxlen=Math.max(maxlen,  indexesa.size());
				int count[]=new int[n];
				for (int x=0;x<indexesa.size();x++)
					count[indexesa.get(x)]++;
				for (int x=1;x<n;x++)
					count[x]=count[x]+count[x-1];
				for (int j=1;j<=26;j++)
				{
					if (i==j || !map.containsKey(j))
						continue;
					
					//now i is the outer number and j is the middle number and both exist in the array and both are different
					
					ArrayList<Integer> indexesb=map.get(j);
					
					for (int p=0;p<indexesb.size();p++)
					{
						int st=indexesb.get(p);
						if (st==0)
							continue;
						for (int q=p;q<indexesb.size();q++)
						{
							int en=indexesb.get(q);
							int bs=q-p+1;
							if (en==n-1)
								continue;
							//System.out.println(i+" "+j+" "+st+" "+en);
							int abefore=count[st-1];
							int aafter=count[n-1]-count[en];
							maxlen=Math.max(maxlen,  bs+2*Math.min(abefore, aafter));
						}
					}
					
				}
			}
			System.out.println(maxlen);
			t--;
		}
	}
}
