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)

# 上位ノードほど @rank が大きい数字となる
class Node
  attr_accessor :parent, :rank

  # 親は自分自身、ランクは 0 として初期化する
  def initialize(n)
    @parent = n
    @rank = 0
  end
end

class UnionFind
  # 0 から n までのノード（つまりn + 1個）を初期化する
  def initialize(n)
    @nodes = (0..n).map { |i| Node.new(i) }
  end

  # 再帰で最も上位の親を探して更新し、その値を返す
  def parent(x)
    return x if @nodes[x].parent == x
    return @nodes[x].parent = parent(@nodes[x].parent)
  end

  # a と b を仲間にする
  def unite(a, b)
    a_parent = parent(a)
    b_parent = parent(b)
    # a 及び b の親が同じ場合は何もしない
    return if a_parent == b_parent

    # a親ランク < b親ランク の場合は、b親をa親の親とする
    # なお、ランクをいじる必要はなし
    if @nodes[a_parent].rank < @nodes[b_parent].rank
      @nodes[a_parent].parent = b_parent
    # その他の場合は、a親をb親の親とする
    else
      @nodes[b_parent].parent = a_parent
    # a親、及びb親のランクが同じ場合は、a親のランクを +1 する
      @nodes[a_parent].rank += 1 if @nodes[a_parent].rank == @nodes[b_parent].rank
    end
  end

  # a と b の親が同じかどうかを返す
  def same?(a, b)
    parent(a) == parent(b)
  end

  # 各ノードの親を全て出力する確認用
  def parents
    @nodes.map(&:parent)
  end
end

n, m = gets.split.map(&:to_i)
edges = []
m.times { edges << gets.split.map(&:to_i) }
sum = 0
edges.each do |edge1|
  tree = UnionFind.new(n)
  edges.each do |edge2|
    if edge1 == edge2
      next
    else
      tree.unite(edge2[0], edge2[1])
    end
  end
  sum += 1 unless tree.same?(edge1[0], edge1[1])
end
puts sum