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: $[\underbrace{a, a, \dots, a}_{x}, \underbrace{b, b, \dots, b}_{y}, \underbrace{a, a, \dots, a}_{x}]$. 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 $t$ ($1 \le t \le 2000$) — the number of test cases. Then $t$ test cases follow.
The first line of the test case contains one integer $n$ ($1 \le n \le 2000$) — the length of $a$. The second line of the test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 26$), 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 $2000$ ($\sum n \le 2000$).
-----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
for _ in range(int(input())):
	n = int(input())
	r = list(map(int, input().split()))
	d = {i: [] for i in range(1, 27)}
	for i in range(n):
		d[r[i]].append(i)
	ans = 1
	for end in range(1, 27):
		c = 1
		while c * 2 <= len(d[end]):
			ll, rr = d[end][c - 1] + 1, d[end][-c] - 1
			dd = [0] * 27
			for p in range(ll, rr + 1):
				dd[r[p]] += 1
			ans = max(ans, max(dd) + 2 * c)
			c += 1
	print (ans)
for _ in range(int(input())):
	n = int(input())
	r = list(map(int, input().split()))
	d = {i: [] for i in range(1, 27)}
	for i in range(n):
		d[r[i]].append(i)
	ans = 1
	for end in range(1, 27):
		c = 1
		while c * 2 <= len(d[end]):
			ll, rr = d[end][c - 1] + 1, d[end][-c] - 1
			dd = [0] * 27
			for p in range(ll, rr + 1):
				dd[r[p]] += 1
			ans = max(ans, max(dd) + 2 * c)
			c += 1
	print (ans)