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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <map>
#include <queue>
#include <list>
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,a,b) for(int i=int(a);i<int(b);++i)
#define crep(i) for(char i='a';i<='z';i++)
#define psortsecond(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.second<b.second;});
#define psortfirst(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.first<b.first;});
#define ALL(x) (x).begin(),(x).end()
int ctoi(const char c){
  if('0' <= c && c <= '9') return (c-'0');
  return -1;
}
using namespace std;
using pii  = pair<int,int>;
long long gcd(long long a, long long b){return (b == 0 ? a : gcd(b, a%b));}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
typedef long long ll;
#define MOD 1000000007
#define EPS 10e-8

ll N,M,A[3000],B[3000],C=0;
vector<vector<ll>> TEN(57);
ll touri[57];

void kentou(ll a,ll b,ll t){
  t++;
  touri[a]=1;
  if(a==b){
    C++;
    return;
  }
  rep(i,TEN[a].size()){
    if(touri[TEN[a][i]]==0 && (t!=1 || TEN[a][i]!=b)){
      kentou(TEN[a][i],b,t);
    }
  }
}

int main(){
  cin >> N >> M;
  rep(i,M){
    cin >> A[i] >> B[i];
    TEN[A[i]].push_back(B[i]);
    TEN[B[i]].push_back(A[i]);
  }
  rep(i,M){
    kentou(A[i],B[i],0);
    rep(j,57){
      touri[j]=0;
    }
  }
  cout << M-C <<  endl;
}