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)

from collections import deque

N,M = map(int,input().split())
sides = [list(map(int,input().split())) for i in range(M)]

vertex = [[] for i in range(N)]
for side in sides:
    vertex[side[0]-1].append(side[1])
    vertex[side[1]-1].append(side[0])

# print(vertex)

ans = 0

for side in sides:
    vertex[side[0]-1].remove(side[1])
    vertex[side[1]-1].remove(side[0])
    # print(vertex)
    for i in range(1,N+1):
        a = 1
        check = deque([i])
        fromcheck = {j:0 for j in range(1,N+1)}
        visit = {i}
        while len(check) != 0:
            v = check.popleft()
            for ver in vertex[v-1]:
                if fromcheck[ver] == 0:
                    fromcheck[ver] = 1
                    check.append(ver)
                    visit.add(ver)
            fromcheck[v] = 1
        if visit != {i for i in range(1,N+1)}:
            a = 0
            break
    if a == 1:
        ans += 1
    vertex[side[0]-1].append(side[1])
    vertex[side[1]-1].append(side[0])

print(M-ans)