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.*;
// scanner.nextInt();
// gatogari.
public class Solution {
    public static void main(String[] args) {       

        Scanner scanner = new Scanner(System.in);
        int test_case = scanner.nextInt();
            
            for(int z = 0; z < test_case; ++z)
            {
                int n = scanner.nextInt();
		int[][] pref = new int[n+1][26];
                int[] array = new int[n];

                for(int i = 0; i < n; ++i)
                    {
                        array[i] = scanner.nextInt()-1;
                    }
                int max = 0;
                
                for(int i = 1; i <= n; ++i)
                {
                    for(int j = 0; j < 26; ++j) pref[i][j] = pref[i-1][j];
                    pref[i][array[i-1]]++;
                    max = Math.max(max, pref[i][array[i-1]]);
                }
                
                //buscar l r por cada subarray
                for(int i = 0; i < n; ++i)
                {
                    for(int j = i; j < n; ++j)
                    {
                        int maxx = 0;
                        for(int k = 0; k < 26; ++k)
                        {
                            int temp =  pref[j+1][k] - pref[i][k];
                            maxx = Math.max(maxx, temp);
                        }
                        int maxxx = 0;
                        for(int k = 0; k < 26; ++k)
                        {
                            int temp =  Math.min(pref[i][k], pref[n][k]-pref[j+1][k]);
                            maxxx = Math.max(maxxx, temp);
                        }
                        max = Math.max(max,maxx+(2*maxxx));
                    }
                }
                System.out.println(max);
                 
                 /*String[] arr = {"a","b","c", "d","e", "f" , "g" , "h", "i" , "j" , "k" , "l", "m", "n", "o", "p", "q" , "r", "s", "t", "u" , "v", "w" ,"x", "y", "z"};
                String arr  = "abcdefghijklmnopqrstuvwxyz";*/
            }
                    
    }
}

