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.parent = [i for i in range(n)]
        self.height = [0]*n

    def get_root(self, i):
        if self.parent[i] == i:
            return i
        else:
            self.parent[i] = self.get_root(self.parent[i])
            return self.parent[i]
        
    def unite(self, i, j):
        root_i = self.get_root(i)
        root_j = self.get_root(j)
        if root_i != root_j:
            if self.height[root_i] < self.height[root_j]:
                self.parent[root_i] = root_j
            else:
                self.parent[root_j] = root_i
                if self.height[root_i] == self.height[root_j]:
                    self.height[root_i] += 1
    
    def is_in_group(self, i, j):
        if self.get_root(i) == self.get_root(j):
            return True
        else:
            return False



def main():
    import sys

    def input(): return sys.stdin.readline().rstrip()

    n, m = map(int, input().split())
    edges = []
    for i in range(m):
        a, b = map(int, input().split())
        edges.append((a-1, b-1))
    
    cnt = 0
    for i in range(m):
        uf = UnionFind(n)
        for j in range(m):
            if i == j: continue
            a, b = edges[j]
            uf.unite(a, b)
        flag = 1
        for j in range(n):
            flag &= uf.is_in_group(0, j)
        if not flag: 
            cnt += 1
    print(cnt)
            
                
if __name__ == '__main__':
    main()