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)

def DFS(v):
  if(already[v]==0):
    already[v]=1
    for i in range(n):
      if(con[v][i]==1):
        if(already[i]==0):
          DFS(i)


n,m=map(int,input().split())
INF=10**15
a=[-1for i in range(m)]
b=[-1for i in range(m)]
count=0
for i in range(m):
  a[i],b[i]=map(int,input().split())
  a[i]-=1
  b[i]-=1
for i in range(m):
  con=[[0 for n in range(n)]for j in range(n)]
  for j in range(n):con[j][j]=1
  already=[0 for i in range(n)]
  for j in range(m):
    if(i!=j):
      con[a[j]][b[j]]=1
      con[b[j]][a[j]]=1

  DFS(0)
  if(0 in already):count+=1
print(count)