# SCC→強連結成分をノードとみなしてDFSで訪問可能判定？

from collections import defaultdict
#from atcoder.scc import SCCGraph

###
# https://github.com/not522/ac-library-python
import sys
import typing


class CSR:
    def __init__(
            self, n: int, edges: typing.List[typing.Tuple[int, int]]) -> None:
        self.start = [0] * (n + 1)
        self.elist = [0] * len(edges)

        for e in edges:
            self.start[e[0] + 1] += 1

        for i in range(1, n + 1):
            self.start[i] += self.start[i - 1]

        counter = self.start.copy()
        for e in edges:
            self.elist[counter[e[0]]] = e[1]
            counter[e[0]] += 1


class SCCGraph:
    '''
    Reference:
    R. Tarjan,
    Depth-First Search and Linear Graph Algorithms
    '''

    def __init__(self, n: int) -> None:
        self._n = n
        self._edges: typing.List[typing.Tuple[int, int]] = []

    def num_vertices(self) -> int:
        return self._n

    def add_edge(self, from_vertex: int, to_vertex: int) -> None:
        self._edges.append((from_vertex, to_vertex))

    def scc_ids(self) -> typing.Tuple[int, typing.List[int]]:
        g = CSR(self._n, self._edges)
        now_ord = 0
        group_num = 0
        visited = []
        low = [0] * self._n
        order = [-1] * self._n
        ids = [0] * self._n

        sys.setrecursionlimit(max(self._n + 1000, sys.getrecursionlimit()))

        def dfs(v: int) -> None:
            nonlocal now_ord
            nonlocal group_num
            nonlocal visited
            nonlocal low
            nonlocal order
            nonlocal ids

            low[v] = now_ord
            order[v] = now_ord
            now_ord += 1
            visited.append(v)
            for i in range(g.start[v], g.start[v + 1]):
                to = g.elist[i]
                if order[to] == -1:
                    dfs(to)
                    low[v] = min(low[v], low[to])
                else:
                    low[v] = min(low[v], order[to])

            if low[v] == order[v]:
                while True:
                    u = visited[-1]
                    visited.pop()
                    order[u] = self._n
                    ids[u] = group_num
                    if u == v:
                        break
                group_num += 1

        for i in range(self._n):
            if order[i] == -1:
                dfs(i)

        for i in range(self._n):
            ids[i] = group_num - 1 - ids[i]

        return group_num, ids

    def scc(self) -> typing.List[typing.List[int]]:
        ids = self.scc_ids()
        group_num = ids[0]
        counts = [0] * group_num
        for x in ids[1]:
            counts[x] += 1
        groups: typing.List[typing.List[int]] = [[] for _ in range(group_num)]
        for i in range(self._n):
            groups[ids[1][i]].append(i)

        return groups

###


N = int(input())
sccgraph = SCCGraph(N)
graph = defaultdict(list)
for i in range(1,N+1):
    A = list(map(int, input().split()))[1:]
    for a in A:
        sccgraph.add_edge(i-1, a-1)
        graph[i-1].append(a-1) 
# 強連結成分を取得
groups = sccgraph.scc()
# 強連結成分を1つのノードとみなして新しいグラフを構築
scc_mapping = [-1] * N
for i, group in enumerate(groups):
    for a in group:
        scc_mapping[a] = i
scc_graph = defaultdict(list)
for i, group in enumerate(groups):
    for a in group:
        a_next = [scc_mapping[n] for n in graph[a]]
        scc_graph[i] = list(set(scc_graph[i]+a_next))

def dfs(start):
    todo = [start,]
    seen = [False] * len(groups)

    while todo:
        tgt = todo.pop()
        if seen[tgt]:
            continue
        seen[tgt] = True

        for next in scc_graph[tgt]:
            todo.append(next)
    
    return seen

connected = dfs(0)      

if(all(connected)):
    print("Yes")
else:
    print("No")