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.Arrays;
import java.util.Scanner;

public class Solution {

    static int freq[][];	// 1 based
    static int A[];

    static void init() {

        freq = new int[A.length + 1][27];
        Arrays.fill(freq[0], 0);
        for(int i = 1; i < A.length; i++)
            for(int k = 1; k <= 26; k++)
                freq[i][k] = freq[i-1][k] + (A[i] == k ? 1 : 0);
    }

    static int getFreq(int i, int j, int k) {
        return freq[j][k] - freq[i-1][k];
    }

    static int solve() {

        int n = A.length - 1;
        int maxLen = 0;
        for(int k = 1; k <= 26; k++)	// take max of 1-palindromes
            maxLen = Math.max(maxLen, getFreq(1, n, k));

        for(int i = 1; i <= n; i++)
            for(int j = i+2; j <= n; j++)
                for(int k = 1; k <= 26; k++) {
                    if(A[i] == A[j]) {
                        int outerSize = 2 * Math.min(getFreq(1, i, A[i]), getFreq(j, n, A[i]));
                        maxLen = Math.max(maxLen, outerSize + getFreq(i+1, j-1, k));
                    }
                }
        return maxLen;
    }

    public static void main(String args[]) {

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0) {
            int n = sc.nextInt();
            A = new int[n + 1];
            for (int i = 1; i <= n; i++)
                A[i] = sc.nextInt();
            init();
            System.out.println(solve());
        }
    }
}

