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)

$n, $m = gets.split.map(&:to_i)
edges = Hash.new {|h, k| h[k] = []}
edge_arr = []
$m.times do
  a, b = gets.split.map {|x| x.to_i - 1}
  edge_arr << [a, b]
  edges[a] << b
  edges[b] << a
end

def dfs(s, edges, visited)
  visited[s] = true
  edges[s].each do |ns|
    dfs(ns, edges, visited) unless visited[ns]
  end
end

def is_union(edges)
  visited = Array.new($n) { false }
  dfs(0, edges, visited)
  visited.all?
end

count = 0
edge_arr.each do |(a, b)|
  edges[a].delete(b)
  edges[b].delete(a)

  count += 1 unless is_union(edges)

  edges[a] << b
  edges[b] << a
end

puts count