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 <algorithm>
#include <iostream>
#include <iomanip>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <numeric>

using namespace std;
using ll = long long;
using ld = long double;
const ll mod =  1000000007;

#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REP_FROM(i, j, n) for (ll i = (j); i < (n); ++i)
#define all(x) (x).begin(),(x).end()

ll power(ll base, ll exponent, ll module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ ll root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
ll inverse(ll x){ return power(x, mod - 2, mod);}
ll gcd(ll a, ll b) {if(a < b) gcd(b, a);ll r;while(r=a%b) {a=b;b=r;}return b;}
struct combination{ vector<ll> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } ll C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

struct UnionFind {
  vector<int> par;

  UnionFind(int n) : par(n, -1) {}
  void init(int n) { par.assign(n, -1); }

  int root(int x) {
    if (par[x] < 0)
      return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y)
      return;
    if (par[x] > par[y])
      swap(x, y);
    par[x] += par[y];
    par[y] = x;
  }

  bool same(int x, int y) {
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }
  int size(int x) { return -par[root(x)]; }
};


signed main() {
  ios::sync_with_stdio(false);
  ll n, m;
  cin >> n >> m;
  UnionFind tree(m+1);
  REP(i ,n) {
    int k, x;
    cin >> k >> x;
    REP(j, k-1) {
      int y;
      cin >> y;
      tree.unite(x, y);
    }
  }
  int root = -1;
  REP(i, n) {
    if(root == -1 && tree.root(1+i) != -1) {
      root = tree.root(1+i);
      continue;
    }
    if(tree.root(1+i) == -1)  continue;
    if(root != tree.root(1+i)) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}