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 PyPy3 (7.3.0) (judge verdict: Accepted)

# coding: utf-8

n, m = list(map(int, input().split(" ")))

class UnionFind:

    def __init__(self, n):
        self.parent = [i for i in range(n)]

    def root(self, n):
        if self.parent[n] == n:
            return n
        else:
            self.parent[n] = self.root(self.parent[n])
            return self.parent[n]

    def union(self, x, y):
        x_root = self.root(x)
        y_root = self.root(y)
        if x_root == y_root:
            pass
        else:
            self.parent[x_root] = y_root

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

ans = 0

input_ = []
for _ in range(m):
    a, b = list(map(int, input().split(" ")))
    a -= 1
    b -= 1
    input_.append((a, b))

for i in range(m):
    uf = UnionFind(n)
    for j in range(m):
        if i == j:
            continue
        a, b = input_[j]
        uf.union(a, b)
    for k in range(1, n):
        if not uf.same(0, k):
            ans += 1
            break
print(ans)