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>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define rrep(i,a,b) for(ll i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
#define MAX_N 100005

int par[MAX_N]; //親
int ran[MAX_N]; //木の深さ

//n要素で初期化
void init(int n){
    for(int i = 0; i < n; i++){
        par[i] = i;
        ran[i] = 0;
    }
}

//木の根を求める
int find(int x){
    if(par[x] == x){
        return x;
    }else{
        return par[x] = find(par[x]);
    }
}

//xとyの属する集合を併合
void unit(int x, int y){
    x = find(x);
    y = find(y);
    if(x == y) return;

    if(ran[x] < ran[y]){
        par[x] = y;
    }else{
        par[y] = x;
        if(ran[x] == ran[y]) ran[x]++;
    }
}

//xとyが同じ集合に属するか否か
bool same(int x, int y){
    return find(x) == find(y);
}

int main(void){
    ll n, m;
    cin >> n >> m;

    ll k[n];
    vector<ll> l[n];

    init(n);

    rep(i, 0, n){
        cin >> k[i];
        rep(j, 0, k[i]){
            ll tmp;
            cin >> tmp;
            l[i].push_back(tmp);
        }

        rep(j, 0, l[i].size() - 1){
            unit(l[i][j], l[i][j + 1]);
        }
    }

    rep(i, 0, n){
        rep(j, 0, l[i].size()){
            if(!same(l[0][0], l[i][j])){
                cout << "NO" << endl;
                return 0;
            }
        }
    }

    cout << "YES" << endl;
    return 0;
}