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 Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner() {
        s = new string[0];
        i = 0;
    }

    public string next() {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int Int() {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0) {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++) {
            Array[i] = Int() + add;
        }
        return Array;
    }

    public long Long() {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0) {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++) {
            Array[i] = Long() + add;
        }
        return Array;
    }
}
class Program {
    static Scanner cin = new Scanner();
    static bool[] visited = new bool[55];
    static bool[,] graph = new bool[55, 55]; //隣接行列
    static int N = cin.Int();
    static int M = cin.Int();
  
    static void Main() {
        var a = new int[M];
        var b = new int[M];

        for (int i = 0; i < M; i++) {
            a[i] = cin.Int();
            b[i] = cin.Int();
            graph[a[i], b[i]] = true;
            graph[b[i], a[i]] = true;
        }

        int ans = 0;
        for (int i = 0; i < M; i++) {
            //i番目の道が存在していないときを考える
            graph[a[i], b[i]] = false;
            graph[b[i], a[i]] = false;

            //すべての頂点を未訪問にする
            for (int j = 1; j <= N; j++) {
                visited[j] = false;
            }

            dfs(1);

            bool bridge = false;
            for (int j = 1; j <= N; j++) {
                if (visited[j] == false) bridge = true; //1つでも訪問できない頂点が合ったらグラフは非連結
            }
            if (bridge) ans++;

            //戻しておく
            graph[a[i], b[i]] = true;
            graph[b[i], a[i]] = true;
        }

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

    static void dfs(int v) {
        visited[v] = true; //頂点vを訪問済みにする
        for (int i = 1; i <= N; i++) {
            if (graph[v, i] == false) continue; //頂点vと頂点iが隣接していなければ飛ばす
            if (visited[i] == true) continue; //頂点iがすでに訪問済みだったら飛ばす
            dfs(i);
        }
    }
}