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.*;
import java.util.ArrayList;
import java.util.List;

public class ThreeBlocksPalindrome {
    static Integer[] dp;
    static int INF = (int) 1e9 + 7;
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int T = Integer.parseInt(in.readLine());
        while(T-- > 0) {
            List<Integer>[] arr = new List[201];
            for(int i = 0; i <= 200; i++) {
                arr[i] = new ArrayList<>();
            }
            int n = Integer.parseInt(in.readLine());
            String[] next = in.readLine().split(" ");
            for(int i = 0; i < n; i++) {
                arr[Integer.parseInt(next[i])].add(i);
            }
            int max = 1;
            for(int i = 1; i <= 200; i++) {
                if(arr[i].isEmpty() || arr[i].size() == 1) {
                    continue;
                }
                for(int j = 1; j <= 200; j++) {
                    if(i == j) {
                        max = Math.max(max, arr[i].size());
                        continue;
                    }
                    List<Integer> list = new ArrayList<>();
                    int x = 0, y = 0;
                    while(x < arr[i].size() && y < arr[j].size()) {
                        if(arr[i].get(x) < arr[j].get(y)) {
                            list.add(0);
                            x++;
                        } else {
                            list.add(1);
                            y++;
                        }
                    }
                    while(x < arr[i].size()) {
                        list.add(0);
                        x++;
                    }
                    while(y < arr[j].size()) {
                        list.add(1);
                        y++;
                    }
                    max = Math.max(max, findLPS(list, y));
                }
            }
            out.println(max);
        }
        out.close();
        in.close();
    }

    private static int findLPS(List<Integer> list, int y) {
        int res = 0;
        int l = 0, r = list.size()-1, x = 0;
        while(l <= r) {
            while(l < r && list.get(l) !=0) {
                l++;
                y--;
            }
            while(l < r && list.get(r) != 0) {
                r--;
                y--;
            }
            if(l >= r) {
                break;
            }
            res = Math.max(2*(++x) + y, res);
            l++; r--;
        }
        return res;
    }
}
