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)

# https://atcoder.jp/contests/abc075/tasks/abc075_c
class UnionFind():
    def __init__(self, N):
        self._root = [-1 for _ in range(N)]

    def find(self, x):
        if self._root[x] < 0: return x
        self._root[x] == self.find(self._root[x])
        return self._root[x]

    def unite(self, x, y):
        gx = self.find(x)
        gy = self.find(y)
        if gx == gy: return False
        if self._root[gx] > self._root[gy]:
            gx, gy = gy, gx
        self._root[gx] += self._root[gy]
        self._root[gy] = gx
        return True

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def size(self, x):
        return -(self._root[self.find(x)])

    def roots(self):
        return [i for i in self._root) if i < 0]


n, m = map(int, input().split())

nodes = []
for _ in range(m):
    a, b = map(int, input().split())
    nodes.append((a, b))

cnt = 0
for i in range(m):
    uf = UnionFind(n)
    for j in range(m):
        if i == j: continue
        uf.unite(nodes[j][0]-1, nodes[j][1]-1)
    cnt += (len(uf.roots()) != 1)
print(cnt)