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)

import copy

def DFS(start, G):
    stack = []
    check = []
    check.append(start)
    stack.append(start)
    while stack:
        node = stack.pop()
        for child in G[node]:    
            if child not in check:
                check.append(child)
                stack.append(child)
    return set(check)

N, M = (int(i) for i in input().split())
G = []
S = []
for i in range(N + 1):
    G.append([])

for i in range(M):
    a,b = (int(j) for j in input().split())
    G[a].append(b)
    G[b].append(a)
    S.append( (a,b) )

res = 0
for side in S:
    a, b = side 
    dashG = copy.deepcopy(G)
    dashG[a].remove(b)
    dashG[b].remove(a)
    if DFS(a, dashG) != DFS(b, dashG):
        res += 1

print(res)