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())

graph = [[False for j in range(N)] for i in range(N)]

am = [0] * M
bm = [0] * M


def judge(visited):
    for v in visited:
        if not v:
            return True
    else:
        # 完走 橋ではない
        return False


def dfs(visited, n):
    visited[n] = True

    for k, v in enumerate(visited):
        if v:
            continue

        if not graph[n][k]:
            continue

        visited = dfs(visited, k)

    return visited


def main():
    for i in range(M):
        # 0-indexed
        ai, bi = map(int, input().split())
        ai -= 1
        bi -= 1

        am[i] = ai
        bm[i] = bi

        graph[ai][bi] = True
        graph[bi][ai] = True

    ans = 0

    for ai, bi in zip(am, bm):
        graph[ai][bi] = False
        graph[bi][ai] = False

        # 訪問済 All False
        visited = dfs([False] * N, 0)

        if judge(visited):
            ans += 1

        # 元に戻す
        graph[ai][bi] = True
        graph[bi][ai] = True

    print(ans)
    return


main()