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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int> P;
P f(int a,int b){
  if(a>b)swap(a,b);
  return P(a,b);
}

int main(){
  int n;cin>>n;
  int N=n*3;
  vector<int> v(N);
  for(int i=0;i<N;i++)cin>>v[i];
  v.push_back(N);
  v.push_back(N);
  set<P> s,t;
  int now=2;
  s.insert(f(v[0],v[1]));
  int ans=0;
  for(int i=0;i<n;i++){
    if(v[now]==v[now+1]&&v[now+1]==v[now+2]){
      ans++;now+=3;
      continue;
    }
    if(v[now]==v[now+2])swap(v[now+1],v[now+2]);
    if(v[now]+1==v[now+2])swap(v[now],v[now+2]);
    bool ok=false;
    if(v[now]==v[now+1])for(P p:s)ok|=(p.first==v[now]||p.second==v[now]);
    else for(P p:s)ok|=(p.first==p.second&&(p.first==v[now]||p.first==v[now+1]||p.first==v[now+2]));
    if(ok){
      ans++;
      if(v[now]==v[now+1]){
        for(P p:s){
          if(p.first==v[now])t.insert(f(p.second,v[now+2]));
          else if(p.second==v[now])t.insert(f(p.first,v[now+2]));
        }
      }
      else{
        for(P p:s){
          if(p.first!=p.second)continue;
          if(p.first==v[now])t.insert(f(v[now+1],v[now+2]));
          if(p.first==v[now+1])t.insert(f(v[now],v[now+2]));
          if(p.first==v[now+2])t.insert(f(v[now+1],v[now]));
        }
      }
    }
    else{
      for(P p:s){
        v[now-1]=p.first;
        v[now-2]=p.second;
        for(int i=now-2;i<now+3;i++)for(int j=now-2;j<i;j++)t.insert(f(v[i],v[j]));
      }
    }
    s.clear();
    swap(s,t);
    now+=3;
  }
  cout<<ans<<endl;
}