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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Linq;

public class Problem_C
{
  public static void Main()
  {
    var inputs = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
    int N = inputs[0], M = inputs[1];

    var from = new int[N];
    var to = new int[N];
    var matrix = new int[N, N];
    for (var i = 0; i < M; i++)
    {
      var edge = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
      from[i] = edge[0] - 1;
      to[i] = edge[1] - 1;
      matrix[from[i], to[i]] = matrix[to[i], from[i]] = 1;
    }

    var count = 0;
    for (var i = 0; i < N; i++)
    {
      // Check how it works when each edge is removed
      matrix[from[i], to[i]] = matrix[to[i], from[i]] = 0;

      var visited = new bool[N];
      Dfs(N, i, matrix, visited);
      if (visited.Contains(false)) count++;

      // Resume original state
      matrix[from[i], to[i]] = matrix[to[i], from[i]] = 1;
    }
    Console.WriteLine($"{count}");
  }

  private static void Dfs(int nodeNum, int node, int[,] matrix, bool[] visited)
  {
    visited[node] = true;
    for (var i = 0; i < nodeNum; i++)
    {
      if (matrix[i, node] == 0) continue;
      if (visited[i]) continue;
      Dfs(nodeNum, i, matrix, visited);
    }
  }
}