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 Ishu
{
    static Scanner scan = new Scanner(System.in);
    static BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out)); 
    static void tc() throws Exception
    {
    int n = scan.nextInt();
    int i,j,k;
    int[] a = new int[n + 2];
    int[][] hash = new int[27][n + 2];
    for(i=1;i<=n;++i)
        {
        a[i] = scan.nextInt();
        for(j=1;j<=26;++j)
            hash[j][i] = hash[j][i-1];
        
        hash[a[i]][i]++;
        }
       
    List<Integer>[] suf = new ArrayList[27];   
    List<Integer>[] pre = new ArrayList[27];
    for(i=0;i<27;++i)
        {
        suf[i] = new ArrayList<Integer>();
        pre[i] = new ArrayList<Integer>();
        }
        
    int[] f = new int[27];
    
    for(i=n;i>=1;--i)    
        {
        f[a[i]]++;
        suf[a[i]].add(i);
        }
    f = new int[27];
    for(i=1;i<=n;++i)
        {
        f[a[i]]++;
        pre[a[i]].add(i);
        }
    
    int ans = 0;
    
    for(i=1;i<=26;++i)    
        {
        for(j=1;j<=f[i];++j)
            {
            int lf = j;
            int rem = f[i] - lf;
            if(rem < lf)
                continue;
                
            int l = pre[i].get(j-1);
            int r = suf[i].get(j-1) - 1;
            int ca = 2 * lf;
            int max = 0;
            for(k=1;k<=26;++k)
                max = Math.max(max, hash[k][r] - hash[k][l]);
                
            ca += max;
            ans = Math.max(ans, ca);    
            }
        }
        
    for(j=1;j<=26;++j)   
        ans = Math.max(ans, hash[j][n]);
    output.write(ans + "\n");
    
    output.flush();
    }
    public static void main(String[] args) throws Exception
    {
    int t = 1;
    t = scan.nextInt();
    while(t-- > 0)
        tc();
    }
}
