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
from sys import stdin,stdout
from math import *
input = stdin.readline
# print=stdout.write
for i in range(int(input())):
	n=int(input())
	ar=list(map(int,input().split()))
	d1=[]
	d2=[]
	for i in range(n):
		d1.append([0]*27)
		d2.append([0]*27)
	d1[0][ar[0]]+=1
	for i in range(1,n):
		for j in range(27):
			d1[i][j]=d1[i-1][j]
		d1[i][ar[i]]+=1
	d2[-1][ar[-1]]+=1
	for i in range(n-2,-1,-1):
		for j in range(27):
			d2[i][j]=d2[i+1][j]
		d2[i][ar[i]]+=1
	ans=1
	# print(d1)
	# print(d2)
	for i in range(n):
		for j in range(i+1,n):
			a=[0]*27
			b=[0]*27
			for k in range(27):
				x=d1[i][k]
				y=d2[j][k]
				a[k]=min(x,y)*2
				b[k]=d1[j-1][k]-d1[i][k]
			if(ans<max(a)+max(b)):
				ans=max(a)+max(b)
	# a=d1[1][26]
	# b=d2[3][26]
	# c=d1[2][14]-d1[1][14]
	# print(a,b,c)
	print(ans)


