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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3

from copy import deepcopy

def dfs(x, adj, visited):
    for y, flag in enumerate(adj[x]):
        if flag and y not in visited:
            visited.add(y)
            dfs(y, adj, visited)

def is_simply_connected(adj):
    n = len(adj)
    visited = {0}
    dfs(0, adj, visited)
    return len(visited) == n

def main():
    n, m = map(int, input().split())
    adj = [[(i == j) for i in range(n)] for j in range(n)]
    edges = set()
    for i in range(m):
        a1, b1 = map(int, input().split())
        a = a1 - 1
        b = b1 - 1
        edges.add((a, b))
        adj[a][b] = True
        adj[b][a] = True

    res = 0
    for a, b in edges:
        adjc = deepcopy(adj)
        adjc[a][b] = False
        adjc[b][a] = False
        if not is_simply_connected(adjc):
            res += 1
    print(res)

main()