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
# Code Forces 634-3: E: Three Blocks Palindrome.
# Mon. Apr. 13, 2020, By: arccosh.

from itertools import combinations

# Popu has odd number of elements!
def solve2(popu):
  cent = sum(popu[::2])
  ret = cent
  lo = 0
  hi = len(popu) - 1
  left = 0
  right = 0
  while hi != lo:
    cent -= popu[lo]
    left += popu[lo + 1]
    lo += 2
    ret = max(ret, cent + 2 * min(left, right))
    while hi != lo and right < left:
      cent -= popu[hi]
      right += popu[hi - 1]
      hi -= 2
      ret = max(ret, cent + 2 * min(left, right))
  return ret

def solve(n, c):
  seen = sorted(set(c))
  if len(seen) == 1:
    return n
  base = [-1] * 256
  prev = [-1] * 256
  link = [n] * n
  for i, v in enumerate(c):
    if base[v] < 0:
      base[v] = i
    else:
      link[prev[v]] = i
    prev[v] = i
  ret = 1
  for a, b in combinations(seen, 2):
    ia = base[a]
    ib = base[b]
    popu = [0]
    while ia < n or ib < n:
      popu.append(0)
      while ia < ib:
        popu[-1] += 1
        ia = link[ia]
      popu.append(0)
      while ib < ia:
        popu[-1] += 1
        ib = link[ib]
    popu.append(0)
    if len(popu) & 1 == 0:
      popu.append(0)
    ret = max(ret, solve2(popu), solve2([0] + popu + [0]))
  return ret

for ca in xrange(int(raw_input())):
  n = int(raw_input())
  c = [int(t) for t in raw_input().split()]
  print('%d' % solve(n, c))
