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;
using System.IO;
using System.Text.RegularExpressions;
using System.Diagnostics;

//var input = Console.ReadLine().Split().Select(int.Parse).ToArray();
namespace AtCoderSolve
{
    
    class Solve
    {
        const int mod = 1000000007;
        
        
        static void Main(string[] args)
        {

            //var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
            //Console.SetOut(sw);

            var input = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int N = input[0];
            int M = input[1];

            var g = new Graph();
            List<List<int>> G = new List<List<int>>();
            bool[] seen = new bool[N];
            g.Gin(G, N);
            g.Sfalse(seen);

            int[,] node = new int[M, 2];
            for(var i = 0; i < M; i++)
            {
                var cin = Console.ReadLine().Split().Select(int.Parse).ToArray();
                node[i, 0] = cin[0] - 1;
                node[i, 1] = cin[1] - 1;
                G[node[i,0]].Add(node[i,1]);
                G[node[i,1]].Add(node[i,0]);
            }

            long ans = 0;

            for(var i = 0; i < M; i++)
            {
                G[node[i, 0]].Remove(node[i, 1]);
                G[node[i, 1]].Remove(node[i, 0]);
                seen[0] = true;
                g.Dfs(G, seen, 0);
                if (!g.AllUnion(seen)) { ans++; }

                g.Sfalse(seen);
                G[node[i, 0]].Add(node[i, 1]);
                G[node[i, 1]].Add(node[i, 0]);
            }

            Console.WriteLine(ans);

            

            //Console.Out.Flush();

        }

        
    }

    public class Calculation
    {

    }

    public class Graph
    {
        public void Dfs(List<List<int>> graph, bool[] seen, int v)
        {
            seen[v] = true;
            //初めて訪れたタイミングでここの処理

            foreach (var next in graph[v])
            {
                if (seen[next] == true) { continue; }
                Dfs(graph, seen, next);
            }
            //行き先がない状態になったらこの部分の処理

        }

        public void Gin(List<List<int>> graph,int size)
        {
            for(var i = 0; i < size; i++)
            {
                graph.Add(new List<int>());
                
            }
        }

        public void Sfalse(bool[] seen)
        {
            for(var i = 0; i < seen.Length; i++)
            {
                seen[i] = false;
            }
        }

        public bool AllUnion(bool[] seen)
        {
            for(var i = 0; i < seen.Length; i++)
            {
                if (!seen[i])
                {
                    return false;
                }
            }
            return true;
        }
    }
}