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 os, sys, io, atexit
sys.stdout = io.BytesIO()
atexit.register(lambda: os.write(1, sys.stdout.getvalue()))
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
write = sys.stdout.write
maxint = float('inf')
 
res = []
T = int(input())
for t in range(T) :
  N = int(input())
  arr = map(int,input().split())

  alpha = list(set(arr))
  C = len(alpha)
 
  pres = {i: [0]*(N+1) for i in alpha }
  postOcc = {i: [N] for i in alpha }
  
  for n in range(N) :
    for c in alpha :
      pres[c][n+1] = pres[c][n] + int(c == arr[n])
    postOcc[arr[N-n-1]].append(N-n-1)

  ans = 0

  #for c in alpha : print c, pres[c], postOcc[c]

  for n in range(N) :
    cc = arr[n]
    mid, out = 0, pres[cc][n+1]
    l, r = n, postOcc[cc][out]
    if l > r : continue
    if l < r :
      for c in alpha : mid = max(mid, pres[c][r]-pres[c][l+1])

    #print "out: ", cc, out
    #print "mid: ", mid, "\tl r: ", l, r
    #print "res: ", out*2 + mid
    #print ""
    ans = max(ans, (out*2 if l < r else (out-1)*2+1) + mid)

  #print ""
  res.append(str(ans))
 
write('\n'.join(res))