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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
#include<tuple>
#include<string>
#include<sstream>
#include<cmath>
#include<climits>
#include<algorithm>
#include<bitset>
#include<set>
#include<stack>
#include<queue>
#include<iomanip>
#include<memory.h>
using namespace std;  
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
#define rep(i,n) for(ll i=0;i<(n);i++)  
#define tii tuple<int,int>
#define tiii tuple<int,int,int>
#define mt make_tuple
#define pb push_back  
#define ALL(a) (a).begin(),(a).end()
#define FST first
#define SEC second  
const int INF = (INT_MAX/2);
const ll LLINF = (LLONG_MAX/2);
const double eps = 1e-5;
const double PI = M_PI;  
#define DEB cerr<<"!"<<endl
#define SHOW(a,b) cerr<<(a)<<" "<<(b)<<endl
#define SHOWARRAY(ar,i,j) REP(a,i)REP(b,j)cerr<<ar[a][b]<<((b==j-1)?((a==i-1)?("\n\n"):("\n")):(" "))
#define DIV 1000007
inline ll pow(ll x,ll n,ll m){ll r=1;while(n>0){if((n&1)==1)r=r*x%m;x=x*x%m;n>>=1;}return r%m;}
#define Nodes 100
class UF{
public:
  int value[Nodes];
  int over[Nodes]; // When over < 0, 'over' keep number of node(using negative number). -1 means this tree is a node.
  int root(int index){
    int t = index;
    stack<int> stack_i;
    while(over[t] >= 0){
      stack_i.push(t);
      t = over[t];
    }
    while(!stack_i.empty()){
      int i = stack_i.top(); stack_i.pop();
      over[i] = t;
    }// reconnect
    return t;
  }
  void merge(int a,int b){
    if(this->root(a) == this->root(b)) return;
    over[root(a)] += over[root(b)]; // over have a number of nodes.
    over[root(b)] = root(a);
  }
  UF(){rep(i,Nodes) over[i] = -1;}
};
int main(){
  int n,m; cin >> n >> m;
  vi va; vi vb;
  rep(i,m){
    int a,b; cin >> a >> b;
    a--,b--;
    va.pb(a); vb.pb(b);
  }
  if(m == 1){cout << 1 << endl; return 0;}
  int ans = 0;
  rep(i,n){
    UF u;
    rep(j,n){
      if(i == j) continue;
      u.merge(va[j],vb[j]);
    }
    rep(i,n){
      if(u.root(0) != u.root(i)){ans++; break;}
    }
  }
  cout << ans << endl;
}