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;

struct UnionFind {
	const int V;
	// par[x] := xのroot
	vector<int> par;
	// sz[x] := xを含む集合のサイズ
	vector<int> sz;
	UnionFind(const int V) :
		V(V),
		par(vector<int>(V)),
		sz(vector<int>(V, 1))
		{
			for(int i = 0; i < V; ++i) par[i] = i;
		}
	bool unite(int x, int y) {
		x = root(x), y = root(y);
		if(same(x, y)) return false;
		if(y < x) swap(x, y);
		par[y] = x;
		sz[x] += sz[y];
		return true;
	}
	int root(int x) {
		if(par[x] == x) return x;
		return (par[x] = root(par[x]));
	}
	bool same(int x, int y) {
		return (root(x) == root(y));
	}
	int size(int x) {
		return sz[root(x)];
	}
};

int main() {
	int n, m; cin >> n >> m;
	vector<int> l[n];
	for(int i = 0; i < n; ++i) {
		int k; cin >> k;
		while(k--) {
			int ll; cin >> ll;
			--ll;
			l[i].push_back(ll);
		}
	}
	UnionFind uf(m);
	for(int i = 0; i < n; ++i) {
		for(int j = 1; j < (int)l[i].size(); ++j) {
			uf.unite(l[i][j - 1], l[i][j]);
		}
	}
	set<int> ans;
	for(int i = 0; i < n; ++i) {
		ans.insert(uf.root(l[i].front()));
	}
	if(ans.size() == 1) {
		cout << "YES" << '\n';
	} else {
		cout << "NO" << '\n';	
	}
	return 0;
}