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: Wrong Answer)

#include<bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define int long long
#define vec(a,n) vector<int> (a)((n))
#define Vec(a,n) vector<string> (a)((n))
#define twovec(a,n,m) vector<vector<int>> a(n,vector<int>(m,0))
#define Twovec(a,n,m) vector<vector<double>> a(n,vector<double>(m,0))
#define P pair<int,int>
#define All(a) (a).begin(),(a).end()
#define Sort(a) sort(All(a))
#define Reverse(a) reverse(All(a))
#define PQ(n) priority_queue<P,vector<P>,greater<P>> (n)
#define pq(n) priority_queue<int> (n)
#define Lower(a,t) lower_bound(All(a),t)
using namespace std;
int max_int = 1000000007;
void Debug(auto a);
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};


signed main(){
    int n,m;
    cin >> n >> m;
    vec(a,m);
    UnionFind tree(m);
    REP(i,n){
        int s,t;
        cin >> t;
        cin >> s;
        a[s]++;
        REP(j,t-1){
            int u;
            cin >> u;
            tree.unite(s,u);
            a[u]++;
        }
    }
    bool ans = true;
    FOR(i,1,m){
        if(tree.same(i-1,i)){
            continue;
        }
        else{
            if(a[i]==0){
                continue;
            }
            ans = false;
        }
    }
    if(ans){
        cout << "YES";
    }
    else{
        cout << "NO";
    }


    return 0;
}



void Debug(auto a){
    cout << "{ ";
    for(auto b: a){
        cout << b << " ";
    }
    cout << "}" << endl;
}