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.*;
public class R634E1 {
    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder out = new StringBuilder();
        int T = Integer.parseInt(in.readLine());
        for (int t = 0; t < T; t++) {
            int N=Integer.parseInt(in.readLine());
            StringTokenizer tok=new StringTokenizer(in.readLine());
            int[] A=new int[N];
            for (int i=0; i<N; i++)
                A[i]=Integer.parseInt(tok.nextToken())-1;
            int max=0;
            for (int a=0; a<26; a++) {
                int $=0;
                for (int c:A)
                    if (c==a)
                        $++;
                max=Math.max(max,$);
                for (int b = 0; b < 26; b++) {
                    List<Integer> s=new ArrayList<>();
                    for (int c:A)
                        if (c==a || c==b)
                            s.add(c);
                    //if (a==0 && b==1) System.out.println(s);
                    int[] sb=new int[s.size()];
                    for (int i=0; i<sb.length; i++)
                        if (s.get(i)==b)
                            sb[i]=1;
                    for (int i=1; i<sb.length; i++)
                        sb[i]+=sb[i-1];
                    for (int x=1, la=0, lb=0, ra=s.size()-1, rb=s.size()-1; x<=s.size()/2; x++, la++, ra--) {
                        while (la<ra && s.get(la)!=a)
                            la++;
                        while (ra>la && s.get(ra)!=a)
                            ra--;
                        if (la>=ra)
                            break;
                        if (la>=lb) {
                            lb=la+1;
                            while (lb<=rb && s.get(lb)!=b)
                                lb++;
                        }
                        if (ra<=rb) {
                            rb=ra-1;
                            while (rb>=lb && s.get(rb)!=b)
                                rb--;
                        }
                        if (lb>rb)
                            break;
                        //System.out.println(la+" "+ra+" "+rb+" "+lb);
                        max=Math.max(max,x+(sb[rb]-(lb>0?sb[lb-1]:0))/*(rb-lb+1)*/+x);
                    }
                }
            }
            out.append(max).append("\n");
        }
        System.out.print(out);
    }
}