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 
try:
	sys.stdin = open('input.txt', 'r') 
	sys.stdout = open('output.txt', 'w')
except:
	pass

def max_freq(l,r):
	d = {}
	for i in range(l,r+1):
		if arr[i] in d:
			d[arr[i]] += 1
		else:
			d[arr[i]] = 1
	return max(d.values()) if len(d)>0 else 0

def palin(st,var):
	i = st
	j = n-1
	match = 0
	stp = -1
	end = -1
	lmax = 0

	while i<j:
		if arr[i]!=var:
			i+=1
		elif arr[j]!=var:
			j-=1
		elif arr[i]==arr[j]==var:
			match += 2
			stp = i
			end = j
			i+=1
			j-=1
			lmax = max(lmax,match+max_freq(i,j))
	return lmax
input = sys.stdin.readline
for tt in range(int(input())):
	n = int(input())
	arr = list(map(int,input().split()))
	if n==1:
		print(1)
		continue
	check = set()
	ans = 0
	for i in range(n):
		if arr[i] not in check:
			check.add(arr[i])
			ans = max(ans,palin(i,arr[i]))
			
	if ans==0:
		print(1)
	else:
		print(ans)
	