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: Wrong Answer)

N, M = gets.split.map(&:to_i)
neighbors = Array.new(N) { Array.new(N) {false} }

edges = []
M.times do
  a, b = gets.split.map(&:to_i)
  edges << [a - 1, b - 1]
  neighbors[a - 1][b - 1] = true
  neighbors[b - 1][a - 1] = true
end

p neighbors

def dfs(v, neighbors, visited)
  # p "visited #{v} #{visited}"
  if visited.all? {|v| v == true }
    return true
  end

  neighbors[v].each_with_index do |u, i|
    if u && visited[i] == false
      visited[i] = true
      # connected
      if dfs(i, neighbors, visited)
        return true
      end
      # visited[i] = false
    end
  end

  return false
end

num = 0
edges.each do |edge|
  # puts "delete #{edge[0]} #{edge[1]}"
  neighbors[edge[0]][edge[1]] = false
  neighbors[edge[1]][edge[0]] = false
  visited = Array.new(N) { false }
  visited[0] = true
  if dfs(0, neighbors, visited) == false
    num += 1
  end
  neighbors[edge[0]][edge[1]] = true
  neighbors[edge[1]][edge[0]] = true
end

puts num