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;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Linq.Expressions;
using System.Text;
class Simple {
    int N, M;
    int[] a, b;
    int[,] G;
    void Solve() {
        //input
        N = io.Int;
        M = io.Int;
        a = new int[M];
        b = new int[M];
        for (int i = 0; i < M;++i){
            a[i] = io.Int-1;
            b[i] = io.Int-1;
        }
        var ans = 0;
        //グラフの隣接行列化
        G = new int[N, N];
        for (int i = 0; i < M;++i){
            G[a[i], b[i]] = 1;
            G[b[i], a[i]] = 1;
        }
        //dfsでi番目の辺を除いたときに連結グラフが成り立つか調べる
        for (int i = 0; i < M;++i){
            G[a[i], b[i]] = 0;
            G[b[i], a[i]] = 0;
            var isVisited = new bool[N];
            isVisited[0] = true;
            ans+=(dfs(0, isVisited)?0:1);
            G[a[i], b[i]] = 1;
            G[b[i], a[i]] = 1;
        }
        //ret
        Console.WriteLine(ans);
    }
    bool dfs(int i,bool[]vis){                
        for (int j = 0; j < N;++j){
            if(G[i,j]==1&&!vis[j]){                
                vis[j] = true;
                dfs(j, vis);
            }
        }
        //連結グラフであるか非連結グラフであるか
        return vis.All(f => f);
    }

    SimpleIO io = new SimpleIO();
    public static void Main(string[] args) { new Simple().Stream(); }
    void Stream() {
        Solve();
        io.writeFlush();
    }
}
class SimpleIO {
    string[] nextBuffer;
    int BufferCnt;
    char[] cs = new char[] { ' ' };
    StreamWriter sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
    public SimpleIO() {
        nextBuffer = new string[0];
        BufferCnt = 0;
        Console.SetOut(sw);
    }
    public string Next() {
        if (BufferCnt < nextBuffer.Length)
            return nextBuffer[BufferCnt++];
        string st = Console.ReadLine();
        while (st == "")
            st = Console.ReadLine();
        nextBuffer = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        BufferCnt = 0;
        return nextBuffer[BufferCnt++];
    }
    public string String => Next();
    public char Char => char.Parse(String);
    public int Int => int.Parse(String);
    public long Long => long.Parse(String);
    public double Double => double.Parse(String);
    public void writeFlush() { Console.Out.Flush(); }
}