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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define foreach(it,v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define chmax(a,b) a=max(a,b)
int n;
int a[6005];
int dp[2005][2005];
void upd(int x,int y,int v){
	chmax(dp[x][y],v);
	chmax(dp[y][x],v);
	chmax(dp[x][n],v);
	chmax(dp[n][x],v);
	chmax(dp[n][y],v);
	chmax(dp[y][n],v);
	chmax(dp[n][n],v);
} 
struct node{
	int x;
	int y;
	int v;
	node(int _x=0,int _y=0,int _v=0):x(_x),y(_y),v(_v){}
};
int main(){
	cin>>n;
	int nn=n*3;
	for(int i=0;i<nn;++i){
		cin>>a[i];
		a[i]--;
	}
	for(int i=0;i<=n;++i){
		for(int j=0;j<=n;++j){
			dp[i][j]=-2e9;
		}
	}
	upd(a[0],a[1],0);
	int cnt=0;
	for(int i=2;i<nn-1;i+=3){
		int x=a[i],y=a[i+1],z=a[i+2];
		if(x==y&&y==z){
			cnt++;
			continue;
		}
		vector<node> v;
		for(int j=0;j<3;++j){
			{
				for(int k=0;k<=n;++k){
					int now=dp[k][n];
					if(y==z){
						chmax(now,dp[k][y]+1);
					}
					v.push_back(node(k,x,now));
				}
			}
			{
				int now=max(dp[n][n],dp[z][z]+1);
				v.push_back(node(x,y,now));
			}
			swap(x,y);
			swap(y,z);
		}
		for(node nd:v){
			int x=nd.x;
			int y=nd.y;
			int v=nd.v;
			upd(x,y,v);
		}
	}
	int ans=dp[n][n];
	int l=a[nn-1];
	chmax(ans,dp[l][l]+1);
	ans+=cnt;
	cout<<ans<<'\n';
	return 0;
}