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: Accepted)

def dfs(v, p)
  @vis[v] = true
  @ord[v] = @k += 1 # 到達時刻
  @low[v] = @ord[v]

  @lines[v]&.each do |to|
    if !@vis[to]

      dfs(to, v)
# 訪問先が閉路の経路であれば現在点もそれに含まれる
      @low[v] = [@low[v], @low[to]].min
# @lowの値が現在と訪問先で等しい
# →現在〜訪問先の経路は閉路に含まれる。
# そうで無いならばその経路は橋である
      @bridge += 1 if @ord[v] < @low[to]
# 訪問先が直前の頂点でなければ→閉路の発生
    elsif to != p
      @low[v] = [@low[v], @ord[to]].min
    end
  end
end
N, M = gets.split.map(&:to_i)
@lines = {}
M.times do
  s, t = gets.split.map(&:to_i)
  (@lines[s-1] ||= []) << t-1
  (@lines[t-1] ||= []) << s-1
end
@vis = Array.new(N, false)# 訪問したか否か
@ord = Array.new(N, 0)# 各頂点の訪問時刻
@low = Array.new(N)
@k = 0 # 訪問時刻
@bridge = 0 # 橋の数
N.times do |i|
  dfs(i, -1) if !@vis[i]
end
puts @bridge