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)

class UnionFind
        attr_accessor :parent, :size

        def initialize(n)
                @parent = (0..n).to_a
                @size = [1] * -~n
        end
	
        def unite(x, y)
                x = root(x)
                y = root(y)
                return if x == y
                if @size[x] < @size[y]
                        @parent[x] = y
                        @size[y] += @size[x]
                else
                        @parent[y] = x
                        @size[x] += @size[y]
                end
                return true
        end

        def root(x)
                while x != @parent[x] do
                        @parent[x], x = @parent[@parent[x]], @parent[x]
                end
                x
        end

        def same?(x, y)
                root(x) == root(y)
        end
end


ab = $<.map { |x| x.split.map(&:to_i) }
N, M = ab.shift


res = M.times.map do |k|
	uf = UnionFind.new(N)
	ab[0...k].each { |x| uf.unite(x[0], x[1]) }
	ab[k+1...N].each { |x| uf.unite(x[0], x[1]) }
	true if	(1..N).map { |k| uf.size[uf.root(k)] }.max == N
end

p res.count(&:nil?)