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)

class UnionFind:

  def __init__(self, n):
    self.par = [i for i in range(n)]
    self.rank = [0] * n
    self.sizes = [1] * n

  def find(self, x):
    if self.par[x] != x:
      self.par[x] = self.find(self.par[x])
    return self.par[x]

  def union(self, x, y):
    x, y = self.find(x), self.find(y)
    if x == y:
      return False
    elif self.rank[x] < self.rank[y]:
      self.par[x] = y
      self.sizes[y] += self.sizes[x]
    elif self.rank[x] > self.rank[y]:
      self.par[y] = x
      self.sizes[x] += self.sizes[y]
    else:
      self.par[y] = x
      self.rank[x] += 1
      self.sizes[x] += self.sizes[y]
    return True

  def same(self, x, y):
    return self.find(x) == self.find(y)

  def size(self, x):
    return self.sizes[self.find(x)]


n, m = list(map(int, input().split()))
edge = []
for i in range(m):
  a, b = list(map(int, input().split()))
  edge.append((a-1, b-1))
ans = 0

for i in range(m):
  uf = UnionFind(n)
  for j in range(m):
    if i==j:
      continue
    uf.union(edge[j][0], edge[j][1])
  ans += uf.union(edge[i][0], edge[i][1])
print(ans)