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.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        StringBuilder sb = new StringBuilder();
        while (t -- > 0){
            int n = sc.nextInt();
            int ans = 0;
            int[] arr = new int[n];
            int[] temp = new int[27];
            for(int i = 0; i < arr.length;i++){
                arr[i] = sc.nextInt();
                temp[arr[i]]++;
            }
            for(int i = 0; i < temp.length;i++){
                if(temp[i] != 0){
                    ans = Math.max(ans,temp[i]);
                }
            }
            for(int i = 0; i < temp.length-1;i++){
                if(temp[i] == 0){
                    continue;
                }
                for(int j = i+1; j  < temp.length;j++){
                    if(temp[j] == 0){
                        continue;
                    }
                    int[] pom = new int[temp[i] + temp[j]];
                    int num = temp[i] + temp[j];
                    int k = 0;
                    for(int x = 0;x < arr.length;x++){
                        if(arr[x] == i || arr[x] == j){
                            pom[k] = arr[x];
                            k++;
                        }
                    }
                    ans = Math.max(ans,resi(pom));
                }
            }
            System.out.println(ans);
        }
    }
    public static void printArray(int[] arr){
        for(int i = 0; i < arr.length;i++){
            System.out.print(arr[i] + " ");
        }
        System.out.println();
    }
    public static int resi(int[] arr){
        int ans = 0;
        int first = arr[0];
        int second = -1;
        for(int i = 1; i < arr.length;i++){
            if(arr[i] != first){
                second = arr[i];
                break;
            }
        }
        int firstOcc = 0;
        int secondOcc = 0;
        for(int i = 0; i < arr.length;i++){
            if(arr[i] == first)firstOcc++;
            if(arr[i] == second)secondOcc++;
        }
        int occFirst = firstOcc;
        int occSecond = secondOcc;
        int pomFirst = second;
        int pomSecond = first;
        int pomOccFirst = secondOcc;
        int pomOccSecond = firstOcc;

        int i = 0;
        int j = arr.length-1;
        int result = 0;
        while (i < j){
            if(arr[i] == first && arr[j] == first){
                result +=2;
                i++;
                j--;
                occFirst -= 2;
                ans = Math.max(ans,result + secondOcc);
            }else if(arr[i] != first){
                i++;
                secondOcc -= 1;
            }else if(arr[j] != first){
                j--;
                secondOcc--;
            }else if(arr[i] != first && arr[j] != first){
                i++;
                j--;
                secondOcc-=2;
            }
        }

        i = 0;
        j = arr.length - 1;
        result = 0;
        while (i < j){
            if(arr[i] == pomFirst && arr[j] == pomFirst){
                result += 2;
                i++;
                j--;
                pomOccFirst -= 2;
                ans = Math.max(ans,pomOccSecond+result);
            }else if(arr[i] != pomFirst){
                i++;
                pomOccSecond -= 1;
            }else if(arr[j] != pomFirst){
                j--;
                pomOccSecond--;
            }else if(arr[i] != pomFirst && arr[j] != pomFirst){
                i++;
                j--;
                pomOccSecond -= 2;
            }
        }

        return ans;
    }
}