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: Time Limit Exceeded)

from collections import deque
n,m = map(int,input().split())
to = [[] for _ in range(n)]
v = []
for _ in range(m):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    to[a].append(b)
    to[b].append(a)
    v.append((a,b))
ans = m
seen = [False]*n
for i in range(m):
    a,b = v[i]
    q = deque()
    q.append((a,-1))
    for j in range(n): seen[j] = False
    seen[a] = True
    flag = True
    while q and flag:
        cur,prev = q.popleft()
        for nxt in to[cur]:
            if nxt == b:
                if cur == a: continue
                ans -= 1
                flag = False
                break
            if nxt == prev: continue
            if seen[nxt]: continue
            q.append((nxt,cur))

print(ans)