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)

n, m = map(int, input().split())
ab = [list(map(int, input().split())) for i in range(m)]
class UnionFind(object):
    def __init__(self, size):
        self.parent = [i for i in range(size)]
        self.rank = [0 for _ in range(size)]
 
 
    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            return self.find(self.parent[x])
 
 
    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
 
        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
        else:
            self.parent[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1
 
    def same(self, x, y):
        return self.find(x) == self.find(y)
def C(x):
    U=UnionFind(n)
    for i in range(m):
        if i==x :
            continue
        U.unite(ab[i][0]-1, ab[i][1]-1)
    for i in range(n):
        for j in range(n):
            if not U.same(i, j):
                return False
    return True
ans=0
for y in range(m):
    if not C(y):
        ans+=1
print(ans)