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>
const int INF = 1e9;
const int MOD = 1e9+7;
const long long LINF = 1e18;
#define dump(x)  cout << 'x' << ' = ' << (x) << ` `;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOREACH(x,a) for(auto& (x) : (a) )
typedef long long ll;
using namespace std;

class union_find {
private:
    vector<ll> par;
    vector<ll> ran;
    vector<ll> m_size;
public:
    union_find(ll n);
    ll find(ll x);
    void unite(ll x, ll y);
    bool same(ll x, ll y);
    ll size(ll x);
};

union_find::union_find(ll n){
    par.resize(n);
    ran.resize(n, 0);
    m_size.resize(n, 1);
    REP(i,n) par[i] = i;
};

ll union_find::find(ll x) {
    if (par[x] == x) return x;
    else return par[x] = find(par[x]);
};

void union_find::unite(ll x, ll y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    ll t = size(x)+size(y);
    m_size[x] = m_size[y] = t;
    if (ran[x] < ran[y]) {
        par[x] = y;
        //size[y] += size[x];
    } else {
        par[y] = x;
        //ize[x] += size[y];
        if (ran[x]==ran[y]) ran[x]++;
    }
};

ll union_find::size(ll x) {
    if (par[x] == x) return m_size[x];
    return size(par[x] = find(par[x]));
}

bool union_find::same(ll x, ll y) {
    return (find(x) == find(y));
};

int main(int argc, char const *argv[]) {
    int n, m;
    cin >> n >> m;
    vector<int> x[m];
    REP(i,n) {
        int k; cin >> k;
        REP(j,k) {
            int l; cin >> l; --l;
            x[l].push_back(i);
        }
    }
    union_find uf(n);
    for (auto &&X : x) {
        if (X.size() <= 1) continue;
        for (int j = 0; j < X.size()-1; ++j) {
            uf.unite(X[j], X[j+1]);
        }
    }
    if (uf.size(0) == n) cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}