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
//package main;

import java.io.*;
import java.util.*;

public final class Main {
    BufferedReader br;
    StringTokenizer stk;
    
    public static void main(String[] args) throws Exception {
        new Main().run();
    }
    
    {
        stk = null;
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    
    long mod = 998244353;
    StringBuilder res = new StringBuilder(100005);
    
    void run() throws Exception {
        int t = ni();
        while(t-- > 0) {
            int n = ni();
            int[] a = new int[n];
            for(int i = 0; i < n; i++) {
                a[i] = ni();
            }
            int[][] prefix = new int[26][n];
            for(int i = 0; i < n; i++) {
                prefix[a[i] - 1][i]++;
            }
            
            int max = 1;
            for(int i = 0; i < 26; i++) {
                for(int j = 1; j < n; j++) {
                    prefix[i][j] += prefix[i][j - 1];
                }
                max = Math.max(max, prefix[i][n - 1]);
            }
            
            for(int i = 1; i < n - 1; i++) {
                for(int j = i; j < n - 1; j++) {
                    int center = 1;
                    for(int k = 0; k < 26; k++) {
                        center = Math.max(center, get(i, j, prefix[k]));
                    }
                    int sides = 0;
                    for(int k = 0; k < 26; k++) {
                        sides = Math.max(sides, Math.min(get(0, i - 1, prefix[k]), get(j + 1, n - 1, prefix[k])));
                    }
                    max = Math.max(max, 2 * sides + center);
                }
            }
            
            res.append(max).append("\n");
        }
        
        System.out.print(res);
    }
    
    //Reader & Writer
    String nextToken() throws Exception {
        if (stk == null || !stk.hasMoreTokens())
            stk = new StringTokenizer(br.readLine(), " ");
        return stk.nextToken();
    }

    String nt() throws Exception {
        return nextToken();
    }

    int ni() throws Exception {
        return Integer.parseInt(nextToken());
    }

    long nl() throws Exception {
        return Long.parseLong(nextToken());
    }
    
    double nd() throws Exception {
        return Double.parseDouble(nextToken());
    }
    
    //Some Misc methods
    int get(int l, int r, int[] a) {
        return l == 0 ? a[r] : a[r] - a[l - 1];
    }
    
    void shuffle(long[] a) {
        Random r = new Random();
        for(int i = 0; i < a.length; i++) {
            int idx = r.nextInt(a.length);
            long temp = a[i];
            a[i] = a[idx];
            a[idx] = temp;
        }
    }
}