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

ans = []
for _ in range(int(input())):
	
	n = int(input())

	a = list(map(int,input().split()))

	f = [[0 for _ in range(26)] for _ in range(n)]

	f[0][a[0]-1] += 1
	for i in range(1,n):
		for j in range(26):
			f[i][j] = f[i-1][j]
		f[i][a[i]-1] += 1


	maxLen = 1
	for i in range(-1,n-1):
		for j in range(i+1,n):
			
			tmp = []
			if i==-1:
				tmp = [0]*26
			else:
				tmp = f[i]

			#print(i,j)
			ar1 = [0]*26
			maxAr2 = 0
			ar3 = [0]*26
			for k in range(26):
				ar1[k] = tmp[k]
				maxAr2 = max(maxAr2,f[j][k] - tmp[k])
				ar3[k] = f[n-1][k] - f[j][k]

			# print(ar1)
			# print(ar3)
			# print(maxAr2)
			# print()
			lenij = 0
			for k in range(26): 
				lenij = max(lenij,min(ar1[k],ar3[k]))
			lenij *= 2
			#print(lenij)
			lenij+=maxAr2
			#print(lenij)
			maxLen = max(maxLen,lenij)

	ans.append(maxLen)
print("\n".join(map(str,ans)))
