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)

#!/usr/bin/python3

n, m = list(map(int, input().split()))
br = []
r = {}
for i in range(m):
    a, b = input().split()
    br.append((a, b))
    try:
        r[a].append(b)
    except KeyError:
        r[a] = []
        r[a].append(b)
    try:
        r[b].append(a)
    except KeyError:
        r[b] = []
        r[b].append(a)

done = {}

def s(node):
    if done[node] == 1:
        return
    done[node] = 1
    for i in r[node]:
        if node in down and i in down:
            continue
        s(i)

def chk():
    for i in r.keys():
        done[i] = 0
    s('1')
    for i in done.keys():
        if done[i] == 0:
            return False
    return True

nbr = 0
for i in br:
    down = i
    if chk() == False:
        nbr += 1

print(nbr)