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: Time Limit Exceeded)

# after YouTube explanation
n = gets.strip.to_i
a = gets.strip.split.map{|e| e.to_i - 1}
dp = Array.new(n+1){Array.new(n+1,-1_000_000_000)}
upd = ->(i,j,x) {
  dp[i][j] = x if dp[i][j] < x
  dp[j][i] = x if dp[j][i] < x
  dp[i][n] = x if dp[i][n] < x
  dp[n][i] = x if dp[n][i] < x
  dp[n][j] = x if dp[n][j] < x
  dp[j][n] = x if dp[j][n] < x
  dp[n][n] = x if dp[n][n] < x
}
upd.call(a[0],a[1],0)

ax = 0
2.step(n*3-3, 3) do |ai|
  x = a[ai]
  y = a[ai+1]
  z = a[ai+2]
  if x==y && y==z
    ax += 1
    next
  end
  q = []
  3.times do
    (n+1).times do |ai| # dp[a][?]
      now = dp[ai][n]
      now = dp[ai][y]+1 if y == z && dp[ai][y]+1 > now
      q.push([ai,x,now])
    end
    now = [dp[n][n], dp[z][z]+1].max  # dp[x][y]
    q.push([x,y,now])
    x,y,z = y,z,x
  end
  q.each do |x,y,now|
    upd.call(x,y,now)
  end
end
ans = dp[n][n]
l = a[-1]
ans = dp[l][l]+1 if dp[l][l]+1> ans
ans += ax
p ans