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.util.*;

public class ProblemE {
    public static void main(String[] args) {
        Scanner scn=new Scanner(System.in);
        int T=scn.nextInt();
        for(int t=0;t<T;t++){
            int N=scn.nextInt();
            int[] A=new int[N];
            for(int i=0;i<N;i++){
                A[i]=scn.nextInt();
            }
            int[][] B= new int[201][N];

            for(int i=0;i<N;i++){
                B[A[i]][i]=A[i];
            }
            int ans=0, acount=0, bcount=0;
            List<Integer> ls;
            int[] a, b;
            for(int i=1;i<201;i++){
                a=B[i];
                for(int j=i+1;j<201;j++){
                    b=B[j];
                    acount=0;
                    bcount=0;
                    ls=new LinkedList<>();
                    for(int k=0;k<N;k++){
                        if(a[k]!=0){
                            ls.add(0);
                            acount++;
                        }
                        if(b[k]!=0){
                            ls.add(1);
                            bcount++;
                        }
                    }
                    if(acount>0&&bcount>0){
                        ans=Math.max(solve(ls, bcount), ans);
                    }
                    ls=new LinkedList<>();
                    acount=0;
                    bcount=0;
                    for(int k=0;k<N;k++){
                        if(a[k]!=0){
                            ls.add(1);
                            acount++;
                        }
                        if(b[k]!=0){
                            ls.add(0);
                            bcount++;
                        }
                    }
                    if(acount>0&&bcount>0){
                        ans=Math.max(solve(ls, acount), ans);
                    }
                    ans=Math.max(ans, acount);
                    ans=Math.max(ans, bcount);
                }
            }
            System.out.println(ans);
        }
    }

    private static int solve(List<Integer> ls, int onesCount) {
        int ans=1;
        int l=0, r=ls.size()-1;
        boolean flag;
        int x=1;
        while(l<=r){
            flag=false;
            while(l<=r){
                if(ls.get(l)==0){
                    flag=true;
                    break;
                }
                else{
                    onesCount--;
                    l++;
                }
            }
            if(flag){
                while(r>=l){
                    if(ls.get(r)==1){
                        onesCount--;
                        r--;
                    }
                    else{
                        if(l==r){
                            ans=Math.max(ans, 1);
                        }
                        else{
                            ans=Math.max(ans, onesCount+2*x);
                        }
                        break;
                    }
                }
            }
            x++;
            l++;
            r--;
        }
        return ans;
    }
}
