Problem:
Score : 600 points

Problem Statement

We have 3N cards arranged in a row from left to right, where each card has an integer between 1 and N (inclusive) written on it. The integer written on the i-th card from the left is A_i.

You will do the following operation N-1 times:

Rearrange the five leftmost cards in any order you like, then remove the three leftmost cards. If the integers written on those three cards are all equal, you gain 1 point.

After these N-1 operations, if the integers written on the remaining three cards are all equal, you will gain 1 additional point.

Find the maximum number of points you can gain.

Constraints

1 \leq N \leq 2000

1 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_{3N}

Output

Print the maximum number of points you can gain.

Sample Input 1

2
1 2 1 2 2 1

Sample Output 1

2

Let us rearrange the five leftmost cards so that the integers written on the six cards will be 2\ 2\ 2\ 1\ 1\ 1 from left to right.

Then, remove the three leftmost cards, all of which have the same integer 2, gaining 1 point.

Now, the integers written on the remaining cards are 1\ 1\ 1.

Since these three cards have the same integer 1, we gain 1 more point.

In this way, we can gain 2 points - which is the maximum possible.

Sample Input 2

3
1 1 2 2 3 3 3 2 1

Sample Output 2

1

Sample Input 3

3
1 1 2 2 2 3 3 3 1

Sample Output 3

3

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

N = gets.to_i
A = gets.split.map(&:to_i)

dp = Array.new(N + 1) { Array.new(N + 1, -1) }
dp[A[0]][A[1]] = 0
idx = 0

add = 0
max = 0
ans = 0

1.upto(N - 1) do |i|
  a, b, c = A[3 * i - 1..3 * i + 1]

  if a == b && b == c
    add += 1
    next
  elsif a == b
    r = a
    q = c

    1.upto(N) do |k|
      if dp[k][r] >= 0 && dp[k][q] < dp[k][r] + 1
        dp[k][q] = dp[k][r] + 1
      end

      if dp[r][k] >= 0 && dp[k][q] < dp[r][k] + 1
        dp[k][q] = dp[r][k] + 1
      end

      ans = dp[k][q] if ans < dp[k][q]
    end
  elsif b == c
    r = b
    q = a

    1.upto(N) do |k|
      if dp[k][r] >= 0 && dp[k][q] < dp[k][r] + 1
        dp[k][q] = dp[k][r] + 1
      end

      if dp[r][k] >= 0 && dp[k][q] < dp[r][k] + 1
        dp[k][q] = dp[r][k] + 1
      end

      ans = dp[k][q] if ans < dp[k][q]
    end
  elsif a == c
    r = a
    q = b

    1.upto(N) do |k|
      if dp[k][r] >= 0 && dp[k][q] < dp[k][r] + 1
        dp[k][q] = dp[k][r] + 1
      end

      if dp[r][k] >= 0 && dp[k][q] < dp[r][k] + 1
        dp[k][q] = dp[r][k] + 1
      end

      ans = dp[k][q] if ans < dp[k][q]
    end
  else
    if dp[c][c] >= 0 && dp[a][b] < dp[c][c] + 1
      dp[a][b] = dp[c][c] + 1
      ans = dp[a][b] if ans < dp[a][b]
    end

    if dp[a][a] >= 0 && dp[b][c] < dp[a][a] + 1
      dp[b][c] = dp[a][a] + 1
      ans = dp[b][c] if ans < dp[b][c]
    end

    if dp[b][b] >= 0 && dp[a][c] < dp[b][b] + 1
      dp[a][c] = dp[b][b] + 1
      ans = dp[a][c] if ans < dp[a][c]
    end
  end

  1.upto(N) do |k|
    dp[k][a] = max if dp[k][a] < max
    dp[k][b] = max if dp[k][b] < max
    dp[k][c] = max if dp[k][c] < max
    dp[a][k] = max if dp[a][k] < max
    dp[b][k] = max if dp[b][k] < max
    dp[c][k] = max if dp[c][k] < max
  end

  dp[a][b] = max if dp[a][b] < max
  dp[b][c] = max if dp[b][c] < max
  dp[a][c] = max if dp[a][c] < max

  max = ans if max < ans
end

l = A.last

if ans < dp[l][l] + 1
  ans = dp[l][l] + 1
end

puts ans + add