Problem: Problem

You were asked by a company to create a program. The company has $ N $ jobs, and each job is numbered from $ 1 $ to $ N $. There are $ M_i $ prerequisite jobs $ X_ {i, j} $ in job $ i $, and work $ X_ {i, j} $ is performed before job $ X_ {i, j} $. And work $ i $ suffers a great loss.

So ask you to find the number of jobs you lost when you did all the jobs in the order that minimized the number of jobs you lost.

Constraints

The input satisfies the following conditions.

* $ 1 \ le N \ le 10 ^ 5 $
* $ 0 \ le M_i \ le N $
* Sum of $ M_i $ does not exceed $ 10 ^ 5 $
* $ 1 \ le X_ {i, j} \ le min (i + 1, N) $

Input

All inputs are given as integers in the following format:


$ N $
$ M_1 $ $ X_ {1,1} $ $ X_ {1,2} $ ... $ X_ {1, M_1} $
$ M_2 $ $ X_ {2,1} $ $ X_ {2,2} $ ... $ X_ {2, M_2} ​​$
::
$ M_N $ $ X_ {N, 1} $ $ X_ {N, 2} $ ... $ X_ {N, M_N} $


The number of jobs $ N $ is given to the $ 1 $ line.
The following $ N $ line gives information on the prerequisite work. Information on work $ i $ is given on the $ i + 1 $ line, separated by blanks. $ M_i $ represents the number of jobs that are the premise of work $ i $, and $ X_ {i, j} $ is the number of jobs that are the premise of job $ i $.

Output

When all the jobs are done in the order that minimizes the number of jobs incurred, the number of jobs incurred at that time is output in one line.

Examples

Input

2
1 1
0


Output

1


Input

3
3 2 1 2
0
1 2


Output

1


Input

5
1 2
1 3
2 1 2
3 1 3 5
0


Output

1

Solution (in CPP):

Solution:
#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*
cout<<fixed<<setprecision(20);
cin.tie(0);
ios::sync_with_stdio(false);
*/
const llint mod=1000000007;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n,i,ans=0,dco=1;cin>>n;
	for(i=1;i<=n;i++){
		int m;cin>>m;
		int aru=0,dme=0;
		while(m--){
			int x;cin>>x;
			if(dme){continue;}
			if(x==i+1){aru=1;continue;}
			if(dco<=x){ans++;aru=0;dme=1;continue;}
		}
		if(aru==0){dco=i+1;}
	}
	cout<<ans<<endl;
	return 0;
}