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)

Edge = Struct.new(:from, :to, :cost)

class PriorityQueue
  def initialize
    @elements = []
  end
  
  def push(n)
    i = @elements.size
    while i > 0
      p = (i-1) / 2
      break if @elements[p].cost <= n.cost
      @elements[i] = @elements[p]
      i = p
    end
    @elements[i] = n
  end
  alias :<< :push
  
  def pop
    min = @elements[0]
    n   = @elements.pop
    unless @elements.empty?
      i = 0
      while i * 2 + 1 < @elements.size
        if @elements[i * 2 + 2].nil? or @elements[i * 2 + 1].cost < @elements[i * 2 + 2].cost
          buf = i * 2 + 1
        else
          buf = i * 2 + 2
        end
        break if @elements[buf].cost >= n.cost
        @elements[i] = @elements[buf]
        i = buf
      end
      @elements[i] = n
    end
    min
  end
  
  def empty?
    @elements.empty?
  end
end

N, M = gets.split.map(&:to_i)

E = Array.new(M)
M.times do
  a, b = gets.split.map(&:to_i)
  a -= 1
  b -= 1
  E << [a, b]
end

sum = 0
E.size.times do |i|
  edges = Hash.new { |h, k| h[k] = [] }
  
  [*0..E.size-1].reject { |x| x == i }.each do |j|
    a, b = E[j]
    edges[a] << Edge.new(a, b, 1)
    edges[b] << Edge.new(b, a, 1)
  end
  
  queue = PriorityQueue.new
  edges[0].each { |e| queue << e }
  
  visit = Array.new(N, false)
  visit[0] = true
  
  dist = Array.new(N, Float::INFINITY)
  dist[0] = 0
  
  until queue.empty?
    e1 = queue.pop
    unless visit[e1.to]
      visit[e1.to] = true
      dist[e1.to] = e1.cost
      edges[e1.to].each { |e2| queue << Edge.new(e2.from, e2.to, e1.cost + e2.cost) }
    end
  end
  sum += 1 if dist.any? { |d| d == Float::INFINITY }
end
puts sum