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: Wrong Answer)

def find_root(x):
    if parent[x] == x:
        return x
    else:
        return find_root(parent[x])


def unite(x, y):
    x = find_root(x)
    y = find_root(y)

    if x != y:
        if rank[x] < rank[y]:
            parent[x] = y
            size[y] += size[x]
        else:
            parent[y] = x
            size[x] += size[y]
            if rank[x] == rank[y]:
                rank[x] += 1


def roots():
    roots = []
    for i in range(N):
        roots.append(find_root(i))
    return roots


def group_count():
    return len(set(roots()))


N, M = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(M)]

parent = [i for i in range(N)]  # 根
rank = [1] * N  # 深さ
size = [1] * N  # iを根とするグループのサイズ

edge = [[ab[M - 1 - i][0] - 1, ab[M - 1 - i][1] - 1] for i in range(M)]

for i in range(M):
    unite(edge[i][0], edge[i][1])
count = group_count()

ans = 0
for j in range(N):
    # 再構築するため、初期化し直す
    parent = [i for i in range(N)]
    rank = [1] * N
    size = [1] * N
    for i in range(M):
        if i == j:
            continue
        unite(edge[i][0], edge[i][1])
    if group_count() > count:
        ans += 1
print(ans)