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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
import collections

sys.setrecursionlimit(10 ** 8)
ni = lambda: int(sys.stdin.readline())
nm = lambda: map(int, sys.stdin.readline().split())
nl = lambda: list(nm())
ns = lambda: sys.stdin.readline().rstrip()


def solve():
    N, M = nm()
    uf = UnionFind(M)
    lang_spoken = set()
    for i in range(N):
        k, *lang = nm()
        assert len(lang) == k
        roots = {uf.find(l - 1) for l in lang}
        lang_spoken |= roots
        roots = list(roots)
        x = roots[0]
        for y in roots[1:]:
            uf.union(x, y)
    return uf.size(0) == len(lang_spoken)


class UnionFind:
    def __init__(self, n):
        # total number of nodes.
        self.n = n
        # node id -> root node id
        self._root_table = list(range(n))
        # root node id -> group size
        self._size_table = [1] * n

    def find(self, x):
        """Returns x's root node id."""
        r = self._root_table[x]
        if r != x:
            # Update the cache on query.
            r = self._root_table[x] = self.find(r)
        return r

    def union(self, x, y):
        """Merges two groups."""
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        # Ensure that x is the larger (or equal) group.
        if self._size_table[x] < self._size_table[y]:
            x, y = y, x

        self._size_table[x] += self._size_table[y]
        self._root_table[y] = x

    def size(self, x):
        return self._size_table[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self._root_table) if x == i]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return "\n".join("{}: {}".format(r, self.members(r)) for r in self.roots())


print(["NO", "YES"][solve()])