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 {
	public static void main (String[] args)throws IOException{
	    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	    int t = Integer.parseInt(br.readLine());
	    StringBuilder sb = new StringBuilder();
	    while(t>0)
	    {
	        int n = Integer.parseInt(br.readLine());
	        String k[] = br.readLine().split(" ");
	        int arr[] = new int[n];
	        int alp[][] = new int[n+1][26];
	        for(int i=0;i<n;i++)
	        {
	            arr[i] = Integer.parseInt(k[i]);
	            alp[i+1][arr[i]-1]++;
	        }
	        for(int i=1;i<=n;i++)
	        {
	            for(int j=0;j<26;j++)
	            {
	                alp[i][j] += alp[i-1][j];
	            }
	        }
	        int ans = 0;
	        for(int i=0;i<26;i++)
	        {
	            ans = Math.max(ans,alp[n][i]);
	        }
	        for(int l=0;l<n;l++)
	        {
	            for(int i=1;l+i<n;i++)
	            {
	                int j = i+l;
	                int val = 0;
	                int out = 0;
	                for(int p=0;p<26;p++)
	                {
	                    val = Math.max(val,alp[j+1][p]-alp[i][p]);
	                    out = Math.max(out,Math.min(alp[i][p],alp[n][p]-alp[j+1][p])*2);
	                }
	                ans = Math.max(ans,out+val);
	            }
	        }
	        sb.append(ans+"\n");
	        t--;
	    }
	    System.out.println(sb);
	}
}