Problem:
Score : 400 points

Problem Statement

On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M.

For CODE FESTIVAL 20XX held on this planet, N participants gathered from all over the planet.

The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}.

Two participants A and B can communicate with each other if and only if one of the following conditions is satisfied:

There exists a language that both A and B can speak.

There exists a participant X that both A and B can communicate with.

Determine whether all N participants can communicate with all other participants.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦K_i≦M

(The sum of all K_i)≦10^5

1≦L_{i,j}≦M

L_{i,1}, L_{i,2}, ..., L_{i,{}K_i} are pairwise distinct.

Partial Score

200 points will be awarded for passing the test set satisfying the following: N≦1000, M≦1000 and (The sum of all K_i)≦1000.

Additional 200 points will be awarded for passing the test set without additional constraints.

Input

The input is given from Standard Input in the following format:

N M
K_1 L_{1,1} L_{1,2} ... L_{1,{}K_1}
K_2 L_{2,1} L_{2,2} ... L_{2,{}K_2}
:
K_N L_{N,1} L_{N,2} ... L_{N,{}K_N}

Output

If all N participants can communicate with all other participants, print YES. Otherwise, print NO.

Sample Input 1

4 6
3 1 2 3
2 4 2
2 4 6
1 6

Sample Output 1

YES

Any two participants can communicate with each other, as follows:

Participants 1 and 2: both can speak language 2.

Participants 2 and 3: both can speak language 4.

Participants 1 and 3: both can communicate with participant 2.

Participants 3 and 4: both can speak language 6.

Participants 2 and 4: both can communicate with participant 3.

Participants 1 and 4: both can communicate with participant 2.

Note that there can be languages spoken by no participant.

Sample Input 2

4 4
2 1 2
2 1 2
1 3
2 4 3

Sample Output 2

NO

For example, participants 1 and 3 cannot communicate with each other.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,o,n) for(long long i = o;i<n;i++)
#define oneforall ios::sync_with_stdio(false);cin.tie(0);
#define all(v) (v).begin(),(v).end()
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define int long long 
const long long inf=1e18;
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
typedef vector<int> vi;
typedef vector<long long> vl;
typedef long long ll;
typedef vector<pair<long, long > > vpll;
typedef vector<pair<int, int > > vpii;
#define FORR(x,arr) for(auto& x:arr)
#define ZERO(a) memset(a,0,sizeof(a))





//3111111111111111111111111111111

	class unionfind {
	vector<int> par, rank;
 
public:
	void init(int n) {
		par.resize(n);
		rank.resize(n);
 
		for (int i = 0; i < n; i++) {
			par[i] = i;
			rank[i] = 0;
		}
	}
 
	int find(int x) {
		if (par[x] == x) return x;
		else return par[x] = find(par[x]);
	}
 
	void unite(int x, int y) {
		x = find(x);
		y = find(y);
		if (x == y) return ;
 
		if (rank[x] < rank[y]) par[x] = y;
		else {
			par[y] = x;
			if (rank[x] == rank[y]) ++rank[x];
		}
	}


	bool same(int x, int y) { return (find(x) == find(y)); }
} uf;


	void solve(){

		ini(n);
		ini(n1);
		uf.init(n+n1);
		FOR(i,0,n){
			ini(x);
			while(x--){
				ini(l);
				l--;
				uf.unite(i,n+l);
			}
		}
		
		set<int> set1;
		FOR(i,0,n){
			set1.insert(uf.find(i));
		}
		if(set1.size() == 1)out("YES");
		else{
			out("NO");
		}







	}	























int32_t main() {
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall
	oneforall 
	oneforall

	solve();
	


	return 0;
}