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

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer sToken = new StringTokenizer(reader.readLine());
        int t = Integer.parseInt(sToken.nextToken());
        StringBuilder ans = new StringBuilder();
        while (t-->0) {
            sToken = new StringTokenizer(reader.readLine());
            int n = Integer.parseInt(sToken.nextToken());
            ArrayList<Integer>[] indexes = new ArrayList[201];
            int[] lengths = new int[201];
            int[][] count = new int[n][201];
            sToken = new StringTokenizer(reader.readLine());
            for (int i=0; i<201; i++) indexes[i] = new ArrayList<Integer>();
            for (int i=0; i<n; i++) {
                int x = Integer.parseInt(sToken.nextToken());
                indexes[x].add(i);
                lengths[x]++;
                if (i==0) { count[i][x]++; continue; }
                count[i] = Arrays.copyOf(count[i-1], 201);
                count[i][x]++;
            }
            int res = Integer.MIN_VALUE;
            for (int i=0; i<201; i++) {
                res = Math.max(res, lengths[i]);
                int length = lengths[i];
                int len = length/2;
                for (int j=0; j<len; j++) {
                    int l = indexes[i].get(j);
                    int r = indexes[i].get(length-j-1);
                    if (l+1==r) continue;
                    int max = 0;
                    for (int k=0; k<201; k++) {
                        max = Math.max(count[r-1][k]-count[l][k], max);
                    }
                    res = Math.max(res, (j+1)*2+max);
                }
            }
            ans.append(res).append("\n");
        }
        System.out.println(ans);
    }
}