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 Main{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		int []a;
		while(t-->0) {
			int n=sc.nextInt();
			a=new int [n];
			for(int i=0;i<n;i++) {
				a[i]=sc.nextInt();
			}
			int [][]cnt=new int [26][n+1];
			//前缀和
			for(int i=0;i<n;i++) {
				for(int j=0;j<26;j++) {
					cnt[j][i+1]=cnt[j][i];
				}
				++cnt[a[i]-1][i+1];
			}
			int ans=0;
			for(int i=0;i<26;i++) ans=Math.max(ans, cnt[i][n-1]);
			//以左右边界为界，枚举左右边界，得出在该分界下，最大值。
			for(int l=0;l<n;l++) {
				for(int r=l;r<n;r++) {
					int cntin=0,cntout=0;
					for(int el=0;el<26;el++) {
						cntin=Math.max(cntin, cnt[el][r+1]-cnt[el][l]);//中间一段
						cntout=Math.max(cntout, Math.min(cnt[el][l], cnt[el][n]-cnt[el][r+1])*2);//旁边两段
					}
					ans=Math.max(ans, cntin+cntout);
				}
			}
			System.out.println(ans);
		}
		
	}
}