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)

N,M = [int(x) for x in str(input()).split()]
E=[]
for i in range(M):
  E.append([int(x) for x in str(input()).split()])


def dfs(u):
  visited[u]=True
  for j in adj[u]:
    if visited[j]==False:
      dfs(j)

count=0
for i in range(M):
  P=E[:i]+E[i+1:]
  visited=[False for x in range(0,N+1)]
  adj=[[0] for k in range(0,N+1)]
  for j in range(1,N+1):
    r=[]
    for l in P:
      if l[0]==j:
        r.append(l[1])
      elif l[1]==j:
        r.append(l[0])
    adj[j]=r
  dfs(1)
  if sum(visited)!=N:
    count+=1
print(count)