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: Runtime Error)

n,m=map(int,input().split())
e=[]
for i in range(n):
    e.append(list(map(int,input().split())))
ans=0
for i in range(m):
    nall=[]
    for j in range(m):
        if i==j:
            continue
        nall.append(e[j])
    visited=[0]*n
    t=[1]
    while len(t)>0:
        tmp=[]
        ntmp=[]
        for p in t:
            visited[p-1]=1
            for q in nall:
                if q[0]==p:
                    if visited[q[1]-1]==0:
                        tmp.append(q[1])
                        visited[q[1]-1]=1
                elif q[1]==p:
                    if visited[q[0]-1]==0:
                        tmp.append(q[0])
                        visited[q[0]-1]=1
                else:
                    ntmp.append(q)
        t=tmp
        nall=ntmp
    if sum(visited)!=n:
        ans+=1
print(ans)