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)

def dfs(v, visited, graph)
  visited[v] = true
  target = graph[v].size.times.select{ |t| graph[v][t] == true }
    if target != []
      target.each{|x| dfs(x, visited, graph) if visited[x] == false }
    end
end

def main()
  n,m = gets.chomp.split(' ').map(&:to_i)
  graph = Array.new(n).map{Array.new(n,false)}
  a = Array.new(n,nil)
  b = Array.new(n,nil)
  ans = 0

  for num in 0..m-1 do
    x,y = gets.chomp.split(' ').map(&:to_i)
    a[num],b[num] = x-1, y-1
    graph[a[num]][b[num]] = true
    graph[b[num]][a[num]] = true
  end

  #solve
  for i in 0..m-1 do
    graph[a[i]][b[i]] = false
    graph[b[i]][a[i]] = false

    visited = Array.new(n,false)
    dfs(0, visited, graph)
    ans = ans + 1 if visited.index(false) != nil

    graph[a[i]][b[i]] = true
    graph[b[i]][a[i]] = true
  end

  puts ans
end

main()