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 (3.8.2) (judge verdict: Accepted)

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

    def find(self, i): # どの集合に属しているか（根ノードの番号）
        if self.parents[i] == i:
            return i
        else:
            self.parents[i] = self.find(self.parents[i]) # 経路圧縮
            return self.parents[i]

    def unite(self, i, j): # 二つの集合を併合
        pi = self.find(i)
        pj = self.find(j)
        if pi != pj:
            if self.rank[pi] < self.rank[pj]:
                self.sizes[pj] += self.sizes[pi]
                self.parents[pi] = pj
            else:
                self.sizes[pi] += self.sizes[pj]
                self.parents[pj] = pi
                if self.rank[pi] == self.rank[pj]:
                    self.rank[pi] += 1
    def same(self, i, j): # 同じ集合に属するかを判定
        return self.find(i)==self.find(j)

    def get_parents(self): # 根ノードの一覧を取得
        for n in range(self.nodes): # findで経路圧縮する
            self.find(n)
        return self.parents

adj = []
N, M = map(int,input().split())
for m in range(M):
    a,b = map(int,input().split())
    adj.append([a-1,b-1])

ans = 0
for i in range(M): # 取り除く辺番号
    uf = UnionFind(N)
    for j in range(M):
        if (i==j): # 辺を追加しない（取り除く）
            continue
        uf.unite(*adj[j])
    
    if len(set(uf.get_parents()))!=1: # 複数の集合にわかれているか確認
        ans += 1
print (ans)