#include <bits/stdc++.h>
using namespace std;
#define N 2005
#define INF 0x3f3f3f3f
#define umx(i,j) ((i)=max(i,j))
//int n,a[N*3],dp2[2],dp1[2][N],dp[2][N][N],add=0;
int n,a[N*3],dp2,dp1[N],dp[N][N],add=0;

inline void upd(int i,int j,int v){
	umx(dp[i][j],v);umx(dp[j][i],v);
	//dp1[i]=max(dp1[i],v);dp1[j]=max(dp1[j],v);
	umx(dp1[i],v);umx(dp1[j],v);
	umx(dp2,v);
}
/*
inline int pnt(int v){
	return v<0?-1:v;
}
*/
struct Node{
	int i,j,v;
	inline Node(int ii,int jj,int vv){i=ii;j=jj;v=vv;}
};
vector<Node> gx;
int main(){
	/*
	freopen("input.in","r",stdin);
	freopen("output.out","w",stdout);
	*/
	scanf("%d",&n);
	for(int i=1;i<=3*n;++i) scanf("%d",a+i);
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j) dp[i][j]=-INF;
		dp1[i]=-INF;
	}
	dp2=0;
	dp1[a[1]]=0;dp1[a[2]]=0;
	dp[a[1]][a[2]]=0;dp[a[2]][a[1]]=0;
	for(int i=1;i<n;++i){
		sort(a+3*i,a+3*i+3);
		if(a[3*i]==a[3*i+2]){
			++add;
			continue;
		}
		gx.push_back(Node(a[3*i+1],a[3*i+2],max(dp2,dp[a[3*i]][a[3*i]]+1)));
		gx.push_back(Node(a[3*i],a[3*i+2],max(dp2,dp[a[3*i+1]][a[3*i+1]]+1)));
		gx.push_back(Node(a[3*i],a[3*i+1],max(dp2,dp[a[3*i+2]][a[3*i+2]]+1)));
		for(int j=1;j<=n;++j){
			gx.push_back(Node(j,a[3*i],max(dp1[j],dp[j][a[3*i+1]]+(a[3*i+1]==a[3*i+2]))));
			gx.push_back(Node(j,a[3*i+1],dp1[j]));
			gx.push_back(Node(j,a[3*i+2],max(dp1[j],dp[j][a[3*i+1]]+(a[3*i+1]==a[3*i]))));
		}
		for(int j=0;j<gx.size();++j) upd(gx[j].i,gx[j].j,gx[j].v);
		gx.clear();
	}
	++dp[a[3*n]][a[3*n]];
	umx(dp1[a[3*n]],dp[a[3*n]][a[3*n]]);
	umx(dp2,dp1[a[3*n]]);
	printf("%d\n",dp2+add);
	return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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 <bits/stdc++.h>
using namespace std;
#define N 2005
#define INF 0x3f3f3f3f
#define umx(i,j) ((i)=max(i,j))
//int n,a[N*3],dp2[2],dp1[2][N],dp[2][N][N],add=0;
int n,a[N*3],dp2,dp1[N],dp[N][N],add=0;

inline void upd(int i,int j,int v){
	umx(dp[i][j],v);umx(dp[j][i],v);
	//dp1[i]=max(dp1[i],v);dp1[j]=max(dp1[j],v);
	umx(dp1[i],v);umx(dp1[j],v);
	umx(dp2,v);
}
/*
inline int pnt(int v){
	return v<0?-1:v;
}
*/
struct Node{
	int i,j,v;
	inline Node(int ii,int jj,int vv){i=ii;j=jj;v=vv;}
};
vector<Node> gx;
int main(){
	/*
	freopen("input.in","r",stdin);
	freopen("output.out","w",stdout);
	*/
	scanf("%d",&n);
	for(int i=1;i<=3*n;++i) scanf("%d",a+i);
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j) dp[i][j]=-INF;
		dp1[i]=-INF;
	}
	dp2=0;
	dp1[a[1]]=0;dp1[a[2]]=0;
	dp[a[1]][a[2]]=0;dp[a[2]][a[1]]=0;
	for(int i=1;i<n;++i){
		sort(a+3*i,a+3*i+3);
		if(a[3*i]==a[3*i+2]){
			++add;
			continue;
		}
		gx.push_back(Node(a[3*i+1],a[3*i+2],max(dp2,dp[a[3*i]][a[3*i]]+1)));
		gx.push_back(Node(a[3*i],a[3*i+2],max(dp2,dp[a[3*i+1]][a[3*i+1]]+1)));
		gx.push_back(Node(a[3*i],a[3*i+1],max(dp2,dp[a[3*i+2]][a[3*i+2]]+1)));
		for(int j=1;j<=n;++j){
			gx.push_back(Node(j,a[3*i],max(dp1[j],dp[j][a[3*i+1]]+(a[3*i+1]==a[3*i+2]))));
			gx.push_back(Node(j,a[3*i+1],dp1[j]));
			gx.push_back(Node(j,a[3*i+2],max(dp1[j],dp[j][a[3*i+1]]+(a[3*i+1]==a[3*i]))));
		}
		for(int j=0;j<gx.size();++j) upd(gx[j].i,gx[j].j,gx[j].v);
		gx.clear();
	}
	++dp[a[3*n]][a[3*n]];
	umx(dp1[a[3*n]],dp[a[3*n]][a[3*n]]);
	umx(dp2,dp1[a[3*n]]);
	printf("%d\n",dp2+add);
	return 0;
}
```