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 FUnionFind
  attr_reader :nodes, :parents, :ranks

  def initialize(n)
    @parents = (0...n).to_a
    @nodes = @parents.map{ |x| [x] }
  end

  def find(i)
    @parents[i]
  end

  def union(i,j)
    i = @parents[i]; j = @parents[j]
    return if i == j
    j,i = i,j if @nodes[i].size < @nodes[j].size
    @nodes[j].each{ |k| @parents[k] = i }
    @nodes[i].concat(@nodes[j])
    @nodes[j].clear
  end
end


N,M = gets.split.map(&:to_i)
edges = (0...M).map{ gets.split.map{ |a| a.to_i-1 } }
p (0...M).count{ |i|
  fs = FUnionFind.new(N)
  (0...M).each{ |j|
    next if i == j
    a,b = edges[j]
    fs.union(a,b)
  }
  a,b = edges[i]
  fs.find(a) != fs.find(b)
}