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 gcd,sqrt,factorial,pi,inf
from collections import deque,defaultdict
from bisect import bisect,bisect_left
from time import time
from itertools import permutations as per
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\r\n')
L=lambda:list(R())
P=lambda x:stdout.write(str(x)+'\n')
lcm=lambda x,y:(x*y)//gcd(x,y)
nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
inv=lambda x:pow(x,N-2,N)
sm=lambda x:(x**2+x)//2
N=10**9+7

def get_ans(x,y):
	global ans
	s=sorted(v[x]+v[y])
	p=[0]*(ln[x])
	q=[0]*(ln[y])
	i=0
	j=0
	cntx=0
	cnty=0
	for k in range(ln[x]+ln[y]):
		if a[s[k]]==x:
			p[i]=cnty
			i+=1
			cntx+=1
		else:
			q[j]=cntx
			j+=1
			cnty+=1
	i=0
	j=0
	cntx=0
	cnty=0
	#print(p,q)
	for k in range(ln[x]+ln[y]):
		if i>=ln[x]//2 and j>=ln[y]//2:break
		if i<ln[x]//2 and a[s[-k-1]]==x:
			ans=max(ans,(i+1)*2+ln[y]-cnty-p[i])
			#print((i+1)*2+ln[y]-cnty-p[i],cnty)
			i+=1
		elif j<ln[y]//2 and a[s[-k-1]]==y:
			ans=max(ans,(j+1)*2+ln[x]-cntx-q[j])
			j+=1
		if a[s[-k-1]]==x:
			cntx+=1
		else:
			cnty+=1

for _ in range(I()):
	n=I()
	a=L()
	g=len(set(a))
	v=[[] for i in range(201)]
	ln=[0]*201
	ans=0
	for i in range(n):
		v[a[i]]+=i,
		ln[a[i]]+=1
		ans=max(ans,ln[a[i]])
	p=list(set(a))
	z=len(p)
	#print(ans)
	for i in range(z-1):
		for j in range(i+1,z):
			get_ans(p[i],p[j])
			#print(ans,p[i],p[j])
	print(ans)