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;

namespace test
{
    class Program
    {
        static void Main(string[] args)
        {
            var input = ReadArrayInt();
            var N = input[0];
            var M = input[1];
            var sides = new int[M, 2];
            for (int i = 0; i < M; i++)
            {
                input = ReadArrayInt();
                sides[i, 0] = input[0];
                sides[i, 1] = input[1];
            }

            var cnt = 0;

            for (int i = 0; i < M; i++)
            {
                var ver = new bool[N + 1];
                ver[0] = true;
                ver[1] = true;
                while (true)
                {
                    var change = false;
                    for (int j = 0; j < M; j++)
                    {
                        if (j != i)
                        {
                            if (!(ver[sides[j, 0]] && ver[sides[j, 1]]) && (ver[sides[j, 0]] || ver[sides[j, 1]]))
                            {
                                ver[sides[j, 0]] = true;
                                ver[sides[j, 1]] = true;
                                change = true;
                            }
                        }
                    }
                    if (!change) break;
                }

                if (!ver.All(b => b)) cnt++;
            }            

            Console.WriteLine(cnt);
            Console.ReadLine();
        }

        static string Read() { return Console.ReadLine(); }
        static int ReadInt() { return int.Parse(Read()); }
        static long ReadLong() { return long.Parse(Read()); }
        static int[] ReadArrayInt() { return Read().Split(' ').Select(s => int.Parse(s)).ToArray(); }
        static long[] ReadArrayLong() { return Read().Split(' ').Select(s => long.Parse(s)).ToArray(); }
    }
}