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: Time Limit Exceeded)

def warshall_floyd(d, n):
    # d: 隣接行列(正方行列)
    # n: 頂点数

    # i -> j の最短距離を計算する
    for k in range(n):
        for i in range(n):
            for j in range(n):
                # i -> k -> j と頂点kを挟む
                d[i][j] = min(d[i][j], d[i][k] + d[k][j])

    return d


# -------------------------------------------------------------------#
import copy


def main():
    # n: 頂点数
    # w: 辺の数
    n, w = map(int, input().split())
    # v -> v ; cost: 0
    d = [[float("Inf")] * n for i in range(n)]
    for i in range(n):
        d[i][i] = 0

    # d[u][v] : 辺u-vのコスト(if not: inf)
    e = []
    for i in range(w):
        # edge: u -> v ; cost : z
        u, v = map(lambda x: int(x) - 1, input().split())
        e.append([u, v])
        # 無向グラフを想定
        d[u][v] = 1
        d[v][u] = 1

    ans = 0
    for (u, v) in e:
        dd = copy.deepcopy(d)
        dd[u][v] = float("Inf")
        dd[v][u] = float("Inf")

        wd = warshall_floyd(dd, n)

        for res in wd:
            if float("Inf") in res:
                ans += 1
                break

    print(ans)


if __name__ == "__main__":
    main()