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 <queue>
#include <vector>
#include <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <functional>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <map>
#include <iomanip>
#include <utility>
using ll=long long;
using ld=long double; 
using namespace std;
//const int INF = 7+(1e+9);
ll GCD(ll x, ll y)
{
    if (y == 0)
    {
        return x;
    }
    else
    {
        return GCD(y, x % y);
    }
}
ll LCM(ll x, ll y)
{
    return x / GCD(x, y) * y;
}
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
ll MOD=1e9+7;

int main(){
    ll n,m;cin>>n>>m;
    vector<pair<ll,ll>> vec(m);
    for(int i=0;i<m;i++){
        int x,y;cin>>x>>y;x--;y--;
        vec[i]=make_pair(x,y);
       // cout<<x<<" "<<y<<endl;
    }
 //   cout<<endl;
    ll c=0;
   // cout<<vec[0].second<<endl;
    for(int i=0;i<m;i++){
        UnionFind tree(n+1);
        for(int j=0;j<m;j++){
            if(i==j){
                
            }else{
                tree.unite(vec[j].first,vec[j].second);
             //   cout<<j<<" "<<vec[j].first<<" "<<vec[j].second<<endl;
            }
          //  cout<<vec[i].first<<" "<<vec
        }
     //   cout<<endl;
       // cout<<endl;
        bool h=true;
        for(int j=1;j<n;j++){
            if(!tree.same(0,j)){
            //    cout<<i<<" "<<j<<endl;
                h=false;
                break;
            }
        }
        if(!h){
            c++;
        }
    }
    cout<<c<<endl;
}