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

def dfs(lst):
    visited=[None]*n
    visited[0]=1
    stack=[]
    for i in lst[0]:
        stack.append(i)
    while stack:
        new=stack.pop()
        if visited[new]==None:
            visited[new]=1
            for i in lst[new]:
                stack.append(i)
    if None in visited:
        return True
    else:
        return False

ans=0
for i in range(m): #橋の回数
    edge=[[] for _ in range(n)]
    for j in range(m): #1個除外する
        if i==j:
            continue
        a=all_edge[j][0]
        b=all_edge[j][1]
        edge[a].append(b)
        edge[b].append(a)
    if dfs(edge):
        ans+=1
print(ans)