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)

class Node
  attr_accessor :parent, :size
  def initialize(n)
    @parent = n
    @size = 1
  end
end

class UnionFindTree
  attr_accessor :nodes
  def initialize(n)
    @nodes = (0..n).map { |i| Node.new(i) }
  end

  def find(x)
    return x if @nodes[x].parent == x
    @nodes[x].parent = find(@nodes[x].parent)
  end

  def unite(a, b)
    pa = find(a)
    pb = find(b)
    return if pa == pb

    if @nodes[pa].size < @nodes[pb].size
      @nodes[pa].parent = pb
    else
      @nodes[pb].parent = pa
    end

    tmp = @nodes[pa].size + @nodes[pb].size
    @nodes[pa].size = tmp
    @nodes[pb].size = tmp

  end

  def same?(a, b)
    find(a) == find(b)
  end

  def size(x)
    return @nodes[find(x)].size
  end

end

n, m = gets.split.map(&:to_i)

edge = []

0.upto(m-1) do |i|
  edge[i] = gets.split.map(&:to_i)
end

ans = 0

#辺iが橋かを判定する
0.upto(m-1) do |i|
  uft = UnionFindTree.new(n)
  0.upto(m-1) do |j|
    next if j==i
    uft.unite(edge[j][0], edge[j][1])
  end
  ans += 1 if uft.size(1) < n
end

puts ans