#include <tuple>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;
const int Maxn=2000;
int n;
int a[Maxn*3+5];
int dp[Maxn+5][Maxn+5];
int row_max[Maxn+5];
int zero,all_max;
vector<tuple<int,int,int> > vis;
void update(int i,int j,int k){
	vis.emplace_back(i,j,k);
	vis.emplace_back(j,i,k);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=3*n;i++){
		scanf("%d",&a[i]);
	}
	memset(row_max,-0x3f,sizeof row_max);
	memset(dp,-0x3f,sizeof dp);
	dp[a[1]][a[2]]=row_max[a[1]]=row_max[a[2]]=zero;
	for(int i=3;i<=3*n;i+=3){
		if(a[i]==a[i+1]&&a[i]==a[i+2]){
			zero--;
			continue;
		}
		for(int _=0;_<3;_++){
			swap(a[i],a[i+1]);
			swap(a[i+1],a[i+2]);
			update(a[i],a[i+1],max(all_max,dp[a[i+2]][a[i+2]]+1));
			for(int j=1;j<=n;j++){
				if(a[i+1]==a[i+2]){
					update(j,a[i],dp[j][a[i+1]]+1);
				}
				update(j,a[i],row_max[j]);
			}
		}
		for(tuple<int,int,int>j:vis){
			int e=get<0>(j),f=get<1>(j),g=get<2>(j);
			dp[e][f]=max(dp[e][f],g);
			row_max[e]=max(row_max[e],g);
			all_max=max(all_max,g);
		}
		vis.clear();
	}
	printf("%d\n",all_max-zero);
	return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

Examples

Input

2
1 2 1 2 2 1


Output

2


Input

3
1 1 2 2 3 3 3 2 1


Output

1


Input

3
1 1 2 2 2 3 3 3 1


Output

3

### Response

```cpp
#include <tuple>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;
const int Maxn=2000;
int n;
int a[Maxn*3+5];
int dp[Maxn+5][Maxn+5];
int row_max[Maxn+5];
int zero,all_max;
vector<tuple<int,int,int> > vis;
void update(int i,int j,int k){
	vis.emplace_back(i,j,k);
	vis.emplace_back(j,i,k);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=3*n;i++){
		scanf("%d",&a[i]);
	}
	memset(row_max,-0x3f,sizeof row_max);
	memset(dp,-0x3f,sizeof dp);
	dp[a[1]][a[2]]=row_max[a[1]]=row_max[a[2]]=zero;
	for(int i=3;i<=3*n;i+=3){
		if(a[i]==a[i+1]&&a[i]==a[i+2]){
			zero--;
			continue;
		}
		for(int _=0;_<3;_++){
			swap(a[i],a[i+1]);
			swap(a[i+1],a[i+2]);
			update(a[i],a[i+1],max(all_max,dp[a[i+2]][a[i+2]]+1));
			for(int j=1;j<=n;j++){
				if(a[i+1]==a[i+2]){
					update(j,a[i],dp[j][a[i+1]]+1);
				}
				update(j,a[i],row_max[j]);
			}
		}
		for(tuple<int,int,int>j:vis){
			int e=get<0>(j),f=get<1>(j),g=get<2>(j);
			dp[e][f]=max(dp[e][f],g);
			row_max[e]=max(row_max[e],g);
			all_max=max(all_max,g);
		}
		vis.clear();
	}
	printf("%d\n",all_max-zero);
	return 0;
}

```