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{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] temp=br.readLine().trim().split(" ");
		
		int numTestCases=Integer.parseInt(temp[0]);
		while(numTestCases-->0){
		    temp=br.readLine().trim().split(" ");
		    int n=Integer.parseInt(temp[0]);
		    int[] arr=new int[n];
		    temp=br.readLine().trim().split(" ");
		    for(int i=0;i<n;i++){
		        arr[i]=Integer.parseInt(temp[i]);
		    }
		    out.println(maxLength(arr));
		}
		
		out.flush();
		out.close();
	}
	public static int maxLength(int[] arr)
	{
	    int n=arr.length;
	    int[][] freqArray=new int[n][201];
	    ArrayList<ArrayList<Integer>> indices=new ArrayList<>();
	    for(int i=0;i<=200;i++){
	        indices.add(new ArrayList<>());
	    }
	    freqArray[0][arr[0]]++;
	    for(int i=1;i<n;i++){
	        for(int j=0;j<=200;j++){
	            freqArray[i][j]=freqArray[i-1][j];
	        }
	        freqArray[i][arr[i]]++;
	    }
	    for(int i=0;i<n;i++){
	        indices.get(arr[i]).add(i);
	    }
	    int maxLength=0;
	    for(int i=0;i<n;i++){
	        for(int j=0;j<=200;j++){
	            maxLength=Math.max(maxLength,freqArray[i][j]);
	        }
	    }
	    for(int i=1;i<=200;i++){
	        int size=indices.get(i).size();
	        if(indices.get(i).size()<2)
	        {
	            continue;
	        }
	        for(int j=0;j<size/2;j++){
	            for(int k=1;k<=200;k++)
	            {
	                int length=2*(j+1)+(freqArray[indices.get(i).get(size-j-1)-1][k]-freqArray[indices.get(i).get(j)][k]);
    	            maxLength=Math.max(maxLength,length);
	            }
	        }
	    }
	    return maxLength;
	}
}