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: Runtime Error)

import sys
sys.setrecursionlimit(10**9+7)
def check(graph,seen,now):
    seen[now] = True 
    for i in range(len(graph[now])):
        if seen[i] :
            continue
        if graph[now][i] == 0 :
            continue
        check(graph,seen,i)
    return seen


#入力
N,M = map(int,input().split())
para = [[i for i in map(int,input().split())] for j in range(M)]
graph =[[0 for i in range(N)] for  j in range(M)]
for i in range(M):
    graph[para[i][0]-1][para[i][1]-1] = 1
    graph[para[i][1]-1][para[i][0]-1] = 1 

#seen = [False]*(N)
ans = 0
for i in range(M):
    seen = [False]*(N)
    graph[para[i][0]-1][para[i][1]-1] = 0
    graph[para[i][1]-1][para[i][0]-1] = 0
    if False in check(graph,seen,0):
        ans += 1
    graph[para[i][0]-1][para[i][1]-1] = 1
    graph[para[i][1]-1][para[i][0]-1] = 1 
print(ans)