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.InputStreamReader;
import java.util.*;
public class E {
	public static void main(String []args) throws Exception
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String input;
		String []splited;
		int q,n,m,t;
		int []arr,arr1; 
		input=br.readLine();
		splited=input.split(" ");
		t=Integer.parseInt(splited[0]);
		int a,b,c,d;
		  
		 int u,v,k;
		 int yy=0;
		 for(int i=0;i<t;i++)
		 {
			 input=br.readLine();
				splited=input.split(" ");
				n=Integer.parseInt(splited[0]);
				arr=new int[n];
				input=br.readLine();
				splited=input.split(" ");
				for(int j=0;j<n;j++)
				{
					arr[j]=Integer.parseInt(splited[j]);
				
				}
				int len=0;
				a=1;b=1;
				a=26;
				int cnt[][]=new int[n][a];
				int c1[]=new int[a];
				ArrayList<Integer> al []=new ArrayList[a];
				for(int j=0;j<a;j++)
				{
					al[j]=new ArrayList<Integer>();
				}
				for(int j=0;j<n;j++)
				{
					al[arr[j]-1].add(j);
				}
				
				for(int j=0;j<n;j++)
				{
					c1[arr[j]-1]++;
					for(int j1=0;j1<a;j1++)
					{
						cnt[j][j1]=c1[j1];
					}
					
				}
				len=1;int ind=0;
				 
				int []cplace=new int[a];
				int []anc=new int[a];
				for(int j=0;j<a;j++)
				{
					cplace[j]=(al[j].size()-1);
				}
				 
				for(int j=0;j<n;j++)
				{
					anc[arr[j]-1]++;
					int c2=cplace[arr[j]-1];
					int max=0;
					cplace[arr[j]-1]--;
					c2=al[arr[j]-1].get(c2);
					if(c2>j)
					{
						
						for(int k1=0;k1<a;k1++)
						{
							if(c2>=1)
							{max=Math.max(max,cnt[c2-1][k1]-cnt[j][k1]);
							}
							
						}
						
						len=Math.max(len,2*anc[arr[j]-1]+max);
						//System.out.println(len);
					}
					
					
				}
				
				System.out.println(len);
				
				
				
				
		 }
	}
}
