def kruskal(dist, N, st):
tree = UnionFind(N)
relation = [[] for i in range(N)]
one_array_dist = list(itertools.chain.from_iterable(dist))
for k in np.argsort(one_array_dist):
i = k % N
j = int(k / N)
if not tree.issame(i, j) and len(relation[i]) < 2\
and len(relation[j]) < 2:
relation[i].append(j)
relation[j].append(i)
tree.unite(i, j)
def go_next():
for i in range(len(relation[next_city])):
if relation[next_city][i] in unvisited_cities:
return relation[next_city][i]
return None
edge = []
for i in range(N):
if len(relation[i]) == 1:
edge.append(i)
relation[edge[0]].append(edge[1])
relation[edge[1]].append(edge[0])
current_city = st
unvisited_cities = set(range(0, N))
tour = [current_city]
unvisited_cities.remove(current_city)
next_city = relation[current_city][0]
while unvisited_cities:
unvisited_cities.remove(next_city)
tour.append(next_city)
next_city = go_next()
return tour
n, m = list(map(int, input().split()))
ab = [list(map(int, input().split())) for i in range(m)]
from unionfind import UnionFind
ans = 0
for i in range(m):
uf = UnionFind(n)
for j, abi in enumerate(ab):
# ai,biの辺を除いたグラフでunionfindを構築
if j == i: continue
a, b = abi
uf.unite(a, b)
a, b = ab[i]
# ai,bi 間の辺を除いた状態で ai,biは同じグループか?(親は存在するか）
# 親がいなければ, 非連結 -> ai,bi間のへんしか存在しないことになる
if not uf.issame(a, b):
ans += 1
print(ans)