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 UnionFind
  def initialize(n)
    @n = n
    @par = Array.new(@n)
    @chl = Hash.new{|h, k| h[k] = []}
    @par.map!.with_index{|x, i| @chl[i]; i}
  end
  def union(a, b)
    pa, pb = [@par[a], @par[b]]
    if pa < pb then
      absorb(a, b)
    elsif pa > pb then
      absorb(b, a)
    end
  end
  def same_group?
    @par.shift
    @par.uniq.size == 1
  end
  def show
    p @par
  end
  
  private
  
  def absorb(pa, ch)
    par = @par[pa]
    chl = @par[ch]
    @chl[chl].each{|c|
	      @chl[par] << c
      @par[c] = par
    }
    @chl.delete(chl)
    @chl[par] << chl
    @par[chl] = par
  end
end

N, M = gets.chomp.split(" ").map(&:to_i)
C = []
M.times{
  C << gets.chomp.split(" ").map(&:to_i)
}

ans = 0
(0...M).each{|i|
  uf = UnionFind.new(N+1)
  C.each_with_index{|(a, b), idx|
    next if idx == i
    uf.union(a, b)
  }
  ans += 1 if not uf.same_group?
}

puts ans