Problem:
Articulation Points

Find articulation points of a given undirected graph G(V, E).

A vertex in an undirected graph is an articulation point (or cut vertex) iff removing it disconnects the graph.

Input

|V| |E|
s0 t0
s1 t1
:
s|E|-1 t|E|-1

, where |V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target verticess of i-th edge (undirected).

Output

A list of articulation points of the graph G ordered by name.

Constraints

1 ≤ |V| ≤ 100,000

0 ≤ |E| ≤ 100,000

The graph is connected

There are no parallel edges

There are no self-loops

Sample Input 1

4 4
0 1
0 2
1 2
2 3

Sample Output 1

2

Sample Input 2

5 4
0 1
1 2
2 3
3 4

Sample Output 2

1
2
3

Solution in Python3 (judge verdict: Wrong Answer)

import sys

f_i = sys.stdin

V, E = map(int, f_i.readline().split())

adj_sets = [set() for i in range(V)]

for l_i in f_i:
    s, t = map(int, l_i.split())
    adj_sets[s].add(t)
    adj_sets[t].add(s)


# dfs traverse

prenum = [0] * V
parent = [0] * V
lowest = [V] * V

import collections

path = collections.deque()
path.append(0)

cnt = 0

while path:
    u = path[-1]
    adj_v = adj_sets[u].copy()
    v_candidate = adj_v.difference(path)
    if v_candidate:
        v = v_candidate.pop()
        cnt += 1
        prenum[v] = cnt
        parent[v] = u
        path.append(v)
        adj_sets[u].remove(v)
        adj_sets[v].remove(u)
    else:
        l = min(prenum[u], lowest[u])
        for v in adj_v:
            l = min(l, prenum[v])
            adj_sets[u].remove(v)
            adj_sets[v].remove(u)
        lowest[u] = l
        lowest[parent[u]] = l
        path.pop()


# output

ans = set()

for u in range(2, V):
    p = parent[u]
    if prenum[p] <= lowest[u]:
        ans.add(p)

if parent.count(0) >= 3:
    ans.add(0)
else:
    ans.discard(0)

ans = list(ans)
ans.sort()

if ans:
    print(*ans, sep = '\n')