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 *

def bfs():
    q = deque([0])
    visited = [False]*N
    visited[0] = True
    
    while q:
        v = q.popleft()
        
        for nv in range(N):
            if G[v][nv]==1:
                if not visited[nv]:
                    visited[nv] = True
                    q.append(nv)
    
    return False in visited
    
N, M = map(int, input().split())
G = [[0]*N for _ in range(N)]
edges = []

for _ in range(M):
    a, b = map(int, input().split())
    G[a-1][b-1] = 1
    G[b-1][a-1] = 1
    edges.append((a-1, b-1))

ans = 0

for a, b in edges:
    G[a][b] = 0
    G[b][a] = 0
    
    if bfs():
        ans += 1
    
    G[a][b] = 1
    G[b][a] = 1

print(ans)