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.data = []
        for i in range(N):
            self.data.append(-1)
    def root(self, a):
        if self.data[a] < 0:
            return a
        else:
            self.data[a] = self.root(self.data[a])
            return self.data[a]
    def same(self, a, b):
        if self.root(a) == self.root(b):
            return 1
        else:
            return 0
    def unite(self, a, b):
        a = self.root(a)
        b = self.root(b)
        if a == b:
            return 0
        if self.data[a] < self.data[b]:
            self.data[a] += self.data[b]
            self.data[b] = a
        else:
            self.data[b] += self.data[a]
            self.data[a] = b
        return 1
s = input().split(" ")
N = int(s[0])
M = int(s[1])
a = []
b = []
for i in range(M):
    s = input().split(" ")
    a.append(int(s[0]) - 1)
    b.append(int(s[1]) - 1)
res = 0
for i in range(M):
    uf = UnionFind(N)
    for j in range(M):
        if j != i:
            uf.unite(a[j], b[j])
    if uf.same(a[i], b[i]) == 0:
        res += 1
print(res)