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>
#define REP(i,n) for(int i=0;i<(n);i++)
using namespace std;

class UFT {
  vector<int> parent;
  vector<int> rank; //木の高さ
  vector<int> size;
public:
  UFT(int n){
    parent = vector<int>(n);
    rank = vector<int>(n,0);
    size = vector<int>(n,1);
    for(int i = 0; i < n; i++)
      parent[i] = i;
  }
  int find_root(int x){
    if ( parent[x] == x ) return x;
    return parent[x] = find_root( parent[x] );
  }
  void unite(int x, int y){
    int u = find_root(x);
    int v = find_root(y);
    if ( u == v ) return;
    if ( rank[u] < rank[v] ){
      parent[u] = v;
      size[v] += size[u];
    }else{
      parent[v] = u;
      size[u] += size[v];
      if ( rank[u] == rank[v] ) rank[x]++;
    }
  }
  bool is_same ( int x, int y ){
    return find_root(x) == find_root(y);
  }
  int size_of ( int x ){
    return size[ find_root(x) ];
  }
};

int N, M;
int K[100000];
set<int> L;
vector<int> l[100000];
int main(){
  cin >> N >> M;
  UFT uft(M);
  REP(i,N){
    cin >> K[i];
    REP(j,K[i]){
      int t;
      cin >> t;
      t--;
      l[i].push_back( t );
      L.insert( t );
      if( j > 0 ){
	uft.unite( l[i][0], l[i][j] );
      }
    }
  }
  if( uft.size_of( l[0][0] ) == L.size() ){
    cout << "YES" << endl;
  }else{
    cout << "NO" << endl;
  }
}