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.Linq;
using System.Text;

class TEST{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		
		int ans = 0;
		for(int i=0;i<M;i++){
			var UF = new UnionFind(N);
			for(int j=0;j<M;j++){
				if(j == i) continue;
				UF.Unite(A[j],B[j]);
			}
			if(!UF.IsUnited(A[i],B[i])) ans++;
		}
		
		Console.WriteLine(ans);
		
	}
	int N,M;
	int[] A,B;
	public Sol(){
		var d = ria();
		N = d[0]; M = d[1];
		A = new int[M];
		B = new int[M];
		for(int i=0;i<M;i++){
			d = ria();
			A[i] = d[0] - 1;
			B[i] = d[1] - 1;
		}
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
}
class UnionFind{
	
	int[] parent;
	int[] mem;
	int compo;
	int N;
	public UnionFind(int n_){
		Initialize(n_);
	}
	
	public void Initialize(int n_){
		N=n_;
		parent=new int[N];
		mem=new int[N];
		for(int i=0;i<N;i++){
			parent[i]=i;
			mem[i]=1;
		}
		compo=N;
	}
	
	public int Parent(int a){
		if(parent[a]==a)return a;
		return parent[a]=Parent(parent[a]);
	}
	
	public bool IsUnited(int a,int b){
		return Parent(a)==Parent(b);
	}
	
	public void Unite(int a,int b){
		a=Parent(a);b=Parent(b);//Parent()を呼ぶことでa以上のノードは全てparentがrootになる
		if(a==b)return;
		parent[a]=b;
		mem[b]+=mem[a];
		compo-=1;
	}
	
	public bool IsRoot(int x){
		return x==parent[x];
	}
	public int MemCnt(int x){
		return mem[Parent(x)];
	}
	
	public void Dump(){
		for(int i=0;i<parent.Length;i++){
			Console.Write(i==0?"{0}":" {0}",parent[i]);
		}
		Console.WriteLine("");
	}
	
	public int Compo{
		get{
			return compo;
		}
	}
	
}