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
import bisect
input=sys.stdin.readline
t=int(input())
for _ in range(t):
  n=int(input())
  a=list(map(int,input().split()))
  cnt_pre=[[] for i in range(27)]# 1-indexed
  sc=[[0]*(n+1) for i in range(27)]# 1-indexed
  for i in range(n):
    cnt_pre[a[i]].append(i+1)
    sc[a[i]][i+1]+=1
  for i in range(27):
    for j in range(1,n+1):
      sc[i][j]+=sc[i][j-1]
  ans=0
  for i in range(27):
    ans=max(ans,sc[i][-1])
  for i in range(27):
    l=len(cnt_pre[i])
    for x in range(1,l//2+1):
      left=cnt_pre[i][x-1]
      right=cnt_pre[i][l-x]
      for j in range(27):
        if i==j:
          continue
        ans=max(ans,2*x+sc[j][right]-sc[j][left-1])
  print(ans)