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.*;
public class Main {

  public static void main(String[] args) { 
      Scanner sc = new Scanner(System.in);
      int t = sc.nextInt();
      while(t-- > 0) {
        int n = sc.nextInt();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++) {
          arr[i] = sc.nextInt();
        }
        int[][] cntLeft = new int[n][27];
        for(int i = 1; i < n; i++) {
          for(int j = 1; j <= 26; j++)
            cntLeft[i][j] = cntLeft[i-1][j];
          int num = arr[i-1];
          cntLeft[i][num]++;
        }

        int[][] cntRight = new int[n][27];
        for(int i = n - 2; i >= 0; i--) {
          for(int j = 1; j <= 26; j++)
            cntRight[i][j] = cntRight[i+1][j];
          int num = arr[i+1];
          cntRight[i][num]++;
        }

        Map<Integer, List<Integer>> idxMap = new HashMap<>();
        for(int i = 1; i <= 26; i++)
          idxMap.put(i, new ArrayList<>());
        for(int i = 0; i < n; i++) {
          idxMap.get(arr[i]).add(i);
        }
        int max = 0;
        for(int num = 1; num <= 26; num++) {
          List<Integer> index = idxMap.get(num);
          for(int a = 0; a < index.size(); a++) {
            int midCnt = 0;
            int idx1 = index.get(a);
            for(int b = a; b < index.size(); b++) {
              int idx2 = index.get(b);
              midCnt++;
              for(int j = 1; j <= 26; j++) {
                int leftCnt = cntLeft[idx1][j];
                int rightCnt = cntRight[idx2][j];
                if(j == num) {
                    leftCnt = 0;
                    rightCnt = 0;
                }
                
                max = Math.max(max, midCnt + Math.min(leftCnt,rightCnt) * 2);
              }
            }
          }
        }
        System.out.println(max);
      }
      
  }
}