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 Python (3.4.3) (judge verdict: Accepted)

class UnionFind():
  def __init__(self,n):
    self.n=n
    self.parents=[-1]*n
  def find(self,x):
    if self.parents[x]<0:
      return x
    else:
      self.parents[x]=self.find(self.parents[x])
      return self.parents[x]
  def union(self,x,y):
    x=self.find(x)
    y=self.find(y)
    if x==y:
      return
    if self.parents[x]>self.parents[y]:
      x,y=y,x
    self.parents[x]+=self.parents[y]
    self.parents[y]=x
  def same(self,x,y):
    return self.find(x)==self.find(y)
  def members(self,x):
    root=self.find(x)
    return [i for i in range(self.n) if self.find(i)==root]
  def roots(self):
    return [i for i,x in enumerate(self.parents) if x<0]
  def group_count(self):
    return len(self.roots())
  def all_group_members(self):
    return {r: self.members(r) for r in self.roots()}
  def __str__(self):
    return '\n'.join('{}: {}'.format(r,self.members(r)) for r in self.roots())
  
n,m=map(int,input().split())
a,b=[0]*m,[0]*m
for i in range(m):
  a[i],b[i]=map(int,input().split())
  a[i]-=1
  b[i]-=1
res=0
for i in range(m):
  UF=UnionFind(n)
  for j in range(m):
    if j!=i:
      UF.union(a[j],b[j])
  if UF.group_count()==2:
    res+=1
print(res)