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 sys
# input = sys.stdin.buffer.readline
def list_input():
	return list(map(int,input().split()))
def mult_input():
	return map(int,input().split())

for nt in range(int(input())):
	n=int(input())
	l=list_input()
	if n==1:
		print (1)
		continue
	elif n==2:
		if l[0]==l[1]:
			print (2)
		else:
			print (1)
		continue
	if len(set(l))==n:
		print (1)
		continue
	counts=[[0 for i in range(n)] for j in range(26)]
	for i in range(1,27):
		count=0
		for j in range(n):
			if l[j]==i:
				count+=1
			counts[i-1][j]=count
	# for i in range(26):
	# 	print (*counts[i])
	ans=0
	for i in range(1,27):
		for j in range(1,27):
			start=0
			end=n-1
			first=0
			while start<end:
				# print (i,j,start,end)
				if l[start]==i and l[end]==i:
					first+=1
					second=(counts[j-1][end-1]-counts[j-1][start])
					# print (i,j,first,second)
					ans=max(first*2+second,ans)
					start+=1
					end-=1
				elif l[start]==i:
					end-=1
				elif l[end]==i:
					start+=1
				else:
					end-=1
					start+=1
	print (ans)

