# 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) 
if N!=1 and len(graph[0])==0:
	print("No")
	exit()
groups = sccgraph.scc()
if 0 not in groups[0]:
    print("No")
    exit()
num_groups = len(groups)
node_to_group = [0] * N

# 各ノードがどの強連結成分に属するかを記録する
for group_index, nodes in enumerate(groups):
    for node in nodes:
        node_to_group[node] = group_index

group_connections = [set() for _ in range(num_groups)]

# グラフ内のノード間の接続を強連結成分間の接続に変換する
for node in range(N):
    current_group = node_to_group[node]
    for neighbor in graph[node]:
        neighbor_group = node_to_group[neighbor]
        if neighbor_group == current_group:
            continue
        group_connections[current_group].add(neighbor_group)

# 各強連結成分が次の強連結成分と直接接続されているかを確認する
for group in range(num_groups - 1):
    if group + 1 not in group_connections[group]:
        print("No")
        exit()
print("Yes")
"🐬😢"