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: Runtime Error)

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

#define MAXN 2001
int n, a[MAXN], mxk[MAXN], mx, tri;

vector<tuple<int,int,int>> ops;
void maxi(int i, int j, int v) {
    ops.emplace_back(i,j,v);
}

void maxip(int i, int j, int v) {
    if (~v) {
        ops.emplace_back(i,j,v+1);
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin>>n;
    vector<vector<int>> dp(n+1, vector<int>(n+1, -1)); 
    fill(mxk, mxk+n+1, -1);
    dp[a[0]][a[1]] = mxk[a[0]] = mxk[a[1]] = 0;
    for (int i = 0; i < 3*n; ++i)cin>>a[i];
    dp[a[0]][a[1]] = 0;
    for (int i = 0; i < n-1; ++i) {
        int p=a[i*3+2],q=a[i*3+3],r=a[i*3+4];
        if (p==q && q==r) {
            tri++;
        } else {
            if(p == q || q == r || p == r){
                if(p == q)q = r;
                else if(q == r)swap(p,q);
                for (int k = 1; k <= n; ++k) {
                    maxip(k, q, dp[k][p]);
                    maxip(k, q, dp[p][k]);
                }
            }
            maxip(q, r, dp[p][p]);
            maxip(p, r, dp[q][q]);
            maxip(p, q, dp[r][r]);
            maxi(q, r, mx);
            maxi(p, r, mx);
            maxi(p, q, mx);
            for (int k = 1; k <= n; ++k) {
                maxi(k, p, mxk[k]);
                maxi(k, q, mxk[k]);
                maxi(k, r, mxk[k]);
            }
            for (auto &o: ops) {
                int i=get<0>(o), j=get<1>(o), v=get<2>(o);
                dp[i][j] = max(dp[i][j], v);
                mxk[i] = max(mxk[i], v);
                mxk[j] = max(mxk[j], v);
                mx = max(mx, v);
            }   
            ops.clear();
        }
    }
    cout<<max(mx, dp[a[n*3-1]][a[n*3-1]] + 1) + tri;
}