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 Ruby (2.3.3) (judge verdict: Wrong Answer)

def dfs(v, current, goal, visited_list = [])
  if current == goal
    @result += 1
    return
  end

  v.each do |w|
    if w[0] == current && !visited_list.include?(w[1])
      dfs(v, w[1], goal, visited_list << current)
    elsif w[1] == current && !visited_list.include?(w[0])
      dfs(v, w[0], goal, visited_list << current)
    end
  end
end
n, m = gets.chomp.split(' ').map(&:to_i)
v = []
m.times { |m| v << gets.chomp.split(' ').map(&:to_i) }
@result = 0

m.times do |m|
  dfs(v.reject.with_index { |e, idx| idx == m }, v[m][0], v[m][1], [])
end

puts m - @result