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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
void chmax(int &a, int b) { a = max(a,b); }

int dp[2005][2005];

int main() {
	// input 
	int n;
	cin >> n;
	int n3 = 3*n;
	vector<int> A(n3);
	rep(i,n3) {cin >> A[i];--A[i];}
	// init
	int m = n + 1;
	rep(i,m) rep(j,m) dp[i][j] = -1e9;
	// init dp
	auto upd = [&](int i, int j, int x) {
		chmax(dp[i][j],x);
		chmax(dp[j][i],x);
		chmax(dp[i][n],x);
		chmax(dp[n][i],x);
		chmax(dp[n][j],x);
		chmax(dp[j][n],x);
		chmax(dp[n][n],x);
	};
	upd(A[0],A[1],0);

	int base = 0;
	for (int ai=2;ai < n3-1;ai+=3) {
		int x=A[ai],y=A[ai+1],z=A[ai+2];
		if (x == y && y==z) {
			base++;
			continue;
		}
		vector<tuple<int,int,int>> q;
		rep(i,3) {
			{
				rep(a,m) {
					int now = dp[a][n];
					if (y == z) chmax(now,dp[a][y]+1);
					q.emplace_back(a,x,now);
				}
			}
			{
				int now = max(dp[n][n],dp[z][z]+1);
				q.emplace_back(x,y,now);
			}
			swap(x,y); swap(y,z);
		}
		for (auto t:q) {
			int i,j,x;
			tie(i,j,x) = t;
			upd(i,j,x);
		}
	}
	int ans = dp[n][n];
	int l =A[n3-1];
	chmax(ans,dp[l][l]+1);
	ans += base;
	cout << ans << endl;
	return 0;
}