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: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;

class Program {
    internal static void Main(string[] args) {
        var inp = ReadLine().Split(' ').Select(s => int.Parse(s)).ToList();
        var N = inp[0];
        var M = inp[1];
        var a = new List<int>();
        var b = new List<int>();
        for (int i = 0; i < M;i++) {
            inp = ReadLine().Split(' ').Select(s => int.Parse(s)).ToList();
            a.Add(inp[0]);
            b.Add(inp[1]);
        }

        var ans = 0;
        for (int i = 1; i <= M;i++) {
            var cnt = a.Count(A => A == i) + b.Count(B => B == i);
            if (cnt == 1) {
                for (int j = 0; j < a.Count;j++) {
                    if (a[j] == i || b[j] == i) {
                        a.RemoveAt(j);
                        b.RemoveAt(j);
                        ans++;
                        i = 1;
                        break;
                    }
                }
            }
        }
        WriteLine(ans);
    }
}