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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

class Solution
{
    static void Main()
    {
        var vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
        int n = vals[0];
        int m = vals[1];
        var graph = Enumerable.Range(0, n + 1).Select(_ => new List<int>()).ToArray();
        for (int i = 0; i < m; i++)
        {
            vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            int a = vals[0];
            int b = vals[1];
            graph[a].Add(b);
            graph[b].Add(a);
        }

        var visited = new HashSet<int>();
        int count = 0;
        foreach (var start in Enumerable.Range(1, n))
        {
            if (visited.Contains(start) || graph[start].Count != 1)
            {
                continue;
            }

            var cur = start;
            visited.Add(cur);
            while (graph[cur].Count == 1 && !visited.Contains(graph[cur].First()))
            {
                var next = graph[cur].First();
                graph[cur].Remove(next);
                graph[next].Remove(cur);
                visited.Add(cur);
                count++;
                cur = next;
            }
        }

        Console.WriteLine(count);
    }
}