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 (3.4.3) (judge verdict: Accepted)

# ABC075C - Bridge
# find the number of bridges in a given undirected connected graph
# bridge := an edge whose removal disconnects a graph
import sys
from copy import deepcopy

input = sys.stdin.readline
sys.setrecursionlimit(10 ** 9)


def dfs(v: int) -> None:
    seen[v] = 1  # make vertex v searched
    for u in g[v]:
        if not seen[u]:
            dfs(u)


def main():  # O(M(N + M))
    global g, seen
    N, M = tuple(map(int, input().split()))
    E = tuple(tuple(map(int, input().split())) for _ in range(M))
    G = [[] for _ in range(N + 1)]  # N+1: 1-idx
    for v, u in E:  # construct a undirected graph
        G[v] += [u]
        G[u] += [v]
    bridges = 0
    for v, u in E:
        g = deepcopy(G)  # subgraph of G except E(v, u)
        g[v].remove(u)
        g[u].remove(v)
        seen = [0] * (N + 1)
        dfs(1)  # search from vertex 1
        bridges += 1 if 0 in seen[1:] else 0
    print(bridges)


if __name__ == "__main__":
    main()