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.io.BufferedInputStream;
import java.io.PrintWriter;
import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintWriter out = new PrintWriter(System.out);
        solve(in, out);
        out.close();
    }

    static void solve(Scanner in, PrintWriter out) {
        int T = in.nextInt();
        for (int t = 0; t < T; t++) {
            int n = in.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) arr[i] = in.nextInt();

            int[][] dp = new int[26+1][n+1];
            for (int i = 0; i < n; i++) {
                dp[arr[i]][i+1] = 1;
            }
            for (int i = 1; i <= dp.length-1; i++) {
                for (int j = 1; j < dp[0].length; j++) {
                    dp[i][j] += dp[i][j-1];
                }
            }
            // fixme: arr pretty !!!
            int ans = 0;
            for (int l = 1; l <= n; l++) {
                for (int r = l; r <= n; r++) {
                    // in
                    int cntIn = 0;
                    for (int k = 1; k <= 26; k++) {
                        cntIn = Math.max(cntIn, dp[k][r] - dp[k][l-1]);
                    }
                    // out
                    int cntOut = 0;
                    for (int k = 1; k <= 26; k++) {
                        cntOut = Math.max(cntOut, Math.min(dp[k][l-1], dp[k][n] - dp[k][r]));
                    }
                    ans = Math.max(ans, cntIn + 2 * cntOut);
                }
            }
            out.println(ans);
        }
    }
}

