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 Div3_634;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class ProblemE {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        int test=Integer.parseInt(br.readLine());
        StringBuilder print=new StringBuilder();
        while(test--!=0){
            int n=Integer.parseInt(br.readLine());
            int a[]=new int[n+1];
            st=new StringTokenizer(br.readLine());
            for(int i=1;i<=n;i++){
                a[i]=Integer.parseInt(st.nextToken());
            }
            int freq[][]=new int[n+1][27];
            for(int i=1;i<=n;i++){
                for(int j=1;j<=26;j++){
                    freq[i][j]=freq[i-1][j];
                }
                freq[i][a[i]]++;
            }
            ArrayList<Integer> ahead[]=new ArrayList[27];
            for(int i=1;i<=26;i++){
                ahead[i]=new ArrayList<>();
            }
            int max=1;
            for(int i=n;i>=1;i--){
                int el=a[i];
                ArrayList<Integer> next=ahead[el];
                int l=freq[i][el];
                for(int j=0;j<next.size();j++){
                    int pos=next.get(j);
                    int r=freq[n][el]-freq[pos-1][el];
                    int b=0;
                    for(int k=1;k<=26;k++){
                        int temp=freq[pos-1][k]-freq[i][k];
                        b=Math.max(b,temp);
                    }
//                    System.out.println(l+" "+b+" "+r+" "+i);
                    max=Math.max(max,b+Math.min(l,r)*2);
                }
                ahead[el].add(i);
            }
            print.append(max+"\n");
        }
        System.out.print(print.toString());
    }
}
