Problem:
Score : 300 points

Problem Statement

You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.

The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.

An edge whose removal disconnects the graph is called a bridge.

Find the number of the edges that are bridges among the M edges.

Notes

A self-loop is an edge i such that a_i=b_i (1 \leq i \leq M).

Double edges are a pair of edges i,j such that a_i=a_j and b_i=b_j (1 \leq i<j \leq M).

An undirected graph is said to be connected when there exists a path between every pair of vertices.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq min(N(N−1)⁄2,50)

1 \leq a_i<b_i \leq N

The given graph does not contain self-loops and double edges.

The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the edges that are bridges among the M edges.

Sample Input 1

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 1

4

The figure below shows the given graph:

The edges shown in red are bridges. There are four of them.

Sample Input 2

3 3
1 2
1 3
2 3

Sample Output 2

0

It is possible that there is no bridge.

Sample Input 3

6 5
1 2
2 3
3 4
4 5
5 6

Sample Output 3

5

It is possible that every edge is a bridge.

Solution in PyPy3 (2.4.0) (judge verdict: Accepted)

from collections import deque


def bfs(NEAR, S, N):  # 幅優先探索  # キュー
    frag = set([S])
    que = deque([S])

    while len(que) > 0:
        q = que.popleft()
        for i in NEAR[q]:
            if i not in frag:
                que.append(i)
                frag.add(i)

    ans = 1
    if len(frag) == N:
        ans = 0
    return ans


n, m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(m)]

near = [set([]) for _ in range(n)]
for i, j in ab:
    near[i - 1].add(j - 1)
    near[j - 1].add(i - 1)

cnt = 0
for i, j in ab:
    near[i - 1].remove(j - 1)
    near[j - 1].remove(i - 1)
    cnt += bfs(near,1,n)
    near[i - 1].add(j - 1)
    near[j - 1].add(i - 1)
print(cnt)