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 {

    Scanner sc = new Scanner(System.in);
    
    public static void main(String[] args) {
        new Main().run();
    }

    void run() {

        int t = sc.nextInt();

        StringBuilder sb = new StringBuilder();

        for (int i=0; i<t; i++) {
            int n = sc.nextInt();
            int[] arr = new int[n];
            int[][] dp = new int[n+1][27];

            for (int j=0; j<n; j++) {
                arr[j] = sc.nextInt();
                dp[j+1][arr[j]]++;
            }

            for (int j=0; j<n; j++) {
                for (int k=1; k<=26; k++) {
                    dp[j+1][k] += dp[j][k];
                }
            }

            int max = 1;

            for (int j=1; j<n; j++) {
                for (int k=j; k<=n; k++) {
                    int cent = 1;
                    for (int l=1; l<=26; l++) {
                        cent = Math.max(cent, dp[k][l]-dp[j-1][l]);
                    }

                    int side = 0;
                    for (int l=1; l<=26; l++) {
                        side = Math.max( side, Math.min(dp[j-1][l], dp[n][l]-dp[k][l]) );
                    }

                    max = Math.max( max, (cent + side * 2) );
                }
            }
            sb.append(max+"\n");

        }

        System.out.println(sb.toString());
    }
}
