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
//189301019.akshay
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.Random;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.HashMap;
import java.util.ArrayList;


public class E1
{ 
    public static void main(String[] args) 
	{ 
		FastReader sc=new FastReader(); 
		StringBuffer ans=new StringBuffer();
		int test=sc.nextInt();
		while(test-->0)
		{
			int n=sc.nextInt();
			int fre[][]=new int[n+1][27];
			HashMap<Integer,ArrayList<Integer>> map =new HashMap<>();
			for(int i =1;i<=n;i++) {
				int temp = sc.nextInt();
				for(int j=1;j<27;j++) {
					if(j == temp)
						fre[i][j]=fre[i-1][j]+1;
					else
						fre[i][j]=fre[i-1][j];
				}
				if(map.containsKey(temp)) {
					map.get(temp).add(i);
				}
				else{
					ArrayList<Integer> t =new ArrayList<>();
					t.add(i);
					map.put(temp,t);
				}
			}
				
			long sum = 1l;
			
			for(int i=1;i<=26;i++) {
				
				if(!map.containsKey(i) || map.get(i).size() == 1)
					continue;
				ArrayList<Integer> occ;
				occ = map.get(i);
				int sp = 0;
				int ep = occ.size()-1;
				while(sp<ep) {
					int s = occ.get(sp);
					int e = occ.get(ep);
					long maxm = fre[e-1][1]-fre[s][1];
					for(int j=1;j<=26;j++) {
						long temp=fre[e-1][j]-fre[s][j];
						if(temp>maxm) {
							maxm = temp;
						}
					}
					if(maxm+((sp+1)*2) > sum) {
						sum=maxm+((sp+1)*2);
					}
						
					++sp;
					--ep;
				}	
			}
			System.out.println(sum);
			
		
		}
		
	    System.out.print(ans);
	} 
    
    static final Random random=new Random();
    
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	
	static void ruffleSort(long[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n);
			long temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	
	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; 
		} 
	} 


} 
