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)

from collections import deque
from copy import deepcopy

N,M = map(int,input().split())
Matrix = [[False for x in range(N)] for j in range(N)]

Path = []
for i in range(M) :
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    Path.append((a,b))
    Matrix[a][b] = True
    Matrix[b][a] = True

INF = 63
ans = 0
for i in range(M) :
    Matrix_duplicated = deepcopy(Matrix)
    a,b = Path[i]
    Matrix_duplicated[a][b] = False
    Matrix_duplicated[b][a] = False
    Distance = [INF for i in range(N)]
    Distance[0] = 0
    Queue = deque([0])
    while Queue :
        now = Queue.popleft()
        Visited.append(now)
        for i in range(N) :
            if Matrix_duplicated[now][i] :
                Matrix_duplicated[now][i] = False
                Matrix_duplicated[i][now] = False
                Distance[i] = min(Distance[now]+1, Distance[i])
                Queue.append(i)
    if INF in Distance :
        ans += 1
print(ans)