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)

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

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        '''
        要素xが属するグループの根を返す
        '''
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
    
    def union(self, x, y):
        '''
        要素xが属するグループと要素yが属するグループとを併合する
        '''
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if self.parents[x] > self.parents[y]:
            x, y = y, x
        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        '''
        要素xが属するグループのサイズ（要素数）を返す
        '''
        return -self.parents[self.find(x)]

    def same(self, x, y):
        '''
        要素x, yが同じグループに属するかどうかを返す
        '''
        return self.find(x) == self.find(y)

    def members(self, x):
        '''
        要素xが属するグループに属する要素をリストで返す
        '''
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        '''
        各要素の親要素の番号を格納するリスト
        '''
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        '''
        グループの数を返す
        '''
        return len(self.roots())

    def all_group_members(self):
        '''
        {ルート要素: [そのグループに含まれる要素のリスト], ...}の辞書を返す
        '''
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

ans = 0
for i in range(M):
    G = UnionFind(N)
    for j in range(M):
        if i == j:
            continue
        a, b = ab[j]
        a, b = a-1, b-1
        G.union(a, b)
    if G.group_count() > 1:
        ans += 1
print(ans)