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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;

/*
<url:https://cf16-final-open.contest.atcoder.jp/tasks/codefestival_2016_final_c>
問題文============================================================
 問題文
 ある星には M 種類の言語があり、1[sim]M の番号が付けられています。
 
 この星のある年のCODE FESTIVALには星中から N 人の参加者が集まりました。
 
 i(1≦i≦N) 人目の参加者は Ki 種類の言語 Li,1,Li,2,…,Li,Ki を話すことが出来ます。
 
 ある 2 人は以下のいずれかの条件を満たすときに限り、コミュニケーションを取ることが出来ます。
 
 2 人ともが話すことの出来る言語が存在する。
 ある人 X が存在して、 2 人ともが X とコミュニケーションを取ることが出来る。
 
 このとき、N 人すべての参加者が他のすべての参加者とコミュニケーションを取ることが出来るかどうかを判定してください。
 
 制約
 2≦N≦105
 1≦M≦105
 1≦Ki≦M
 Kiの総和≦105
 1≦Li,j≦M
 Li,1,Li,2,…,Li,Ki は相異なる。
=================================================================

解説=============================================================
UnionFind
 
 問題条件より、結果的に参加者達が全域木のように結ばれればいいということが分かる
 よってunion Find を用いて、コミュニケーション可能か集団の集合を構築し、
 その集合の数によって判定すれば良い
================================================================
*/

struct UnionFind {
    vector<int> data;
    UnionFind(int size) : data(size, -1) { }
    bool unionSet(int x, int y) {
        x = root(x); y = root(y);
        if (x != y) {
            if (data[y] < data[x]) swap(x, y);
            data[x] += data[y]; data[y] = x;
        }
        return x != y;
    }
    bool findSet(int x, int y) {
        return root(x) == root(y);
    }
    int root(int x) {
        return data[x] < 0 ? x : data[x] = root(data[x]);
    }
    int size(int x) {
        return -data[root(x)];
    }
};
int main(void) {
	cin.tie(0); ios::sync_with_stdio(false);
    ll N,M; cin >> N >> M;
    vector<vector<ll>> lang(M);
    for(int i = 0; i < N;i++){
        ll K; cin >> K;
        for(int j = 0; j < K;j++){
            ll L; cin >> L;
            L--; lang[L].push_back(i);
        }
    }
    UnionFind uf(N);
    for(int i = 0; i < M;i++){
        for(int j = 0; j < (int)lang[i].size() - 1;j++){
            uf.unionSet(lang[i][j],lang[i][j+1]);
        }
    }
    set<ll> s;
    for(int i = 0; i < N;i++){
        s.insert(uf.root(i));
    }
    if(s.size() == 1){
        cout << "YES" << endl;
    }else{
        cout << "NO" << endl;
    }
	return 0;
}