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
t=int(input())
for i in range(t):
  n=int(input())
  a=list(map(int, input().split(" ")))
  u={k:[0 for i in range(n)] for k in range(1, 27)}
  u[a[0]][0]=1
  for k in range(1, n):
    for pro in range(1, 27):
      u[pro][k]=u[pro][k-1]
    u[a[k]][k]=u[a[k]][k-1]+1
  maxx=0
  for pro in range(1, 27):
    cnt=0
    i=0
    j=n-1
    lmax=0
    while(i<n and a[i]!=pro):
      i+=1
    while(j>-1 and a[j]!=pro):
      j-=1
    if(i==j):
      maxx=max(maxx, 1)
    flag=False
    while(i<j):
      if(a[i]==pro):
        while(a[j]!=pro):
          j-=1
          if(j==i):
            lmax=max(2*(cnt)+1, lmax)
            flag=True
            break
      if flag:
        break
      flag=False
      if(a[j]==pro):
        while(a[i]!=pro):
          i+=1
          if(j==i):
            lmax=max(2*(cnt)+1, lmax)
            flag=True
            break
      if flag:
        break
      if(a[i]==pro==a[j]):
        cnt+=1
      mex=0
      for tro in range(1, 27):
       if tro!=pro:
        mex=max(mex, u[tro][j-1]-u[tro][i])
      lmax=max(mex+2*cnt, lmax)
      i+=1
      j-=1
      if(i==j and a[i]==pro):
        lmax=max(lmax, 2*cnt+1)
    maxx=max(lmax, maxx)
  print(maxx)