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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < n; ++i)
#define ALL(c) (c).begin(), (c).end()
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())
using namespace std;

using ll = long long;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline void dump(vector<T> v) { for (auto& x : v) cerr << x << " "; cerr << endl; }
template<class T> inline void dump(vector<pair<T, T>> v) { for (auto& p : v) cerr << p.first << " " << p.second << endl; }
template<class T> inline void dump(vector<vector<T>> vv) { for (auto& v : vv) {for (auto& x : v) cerr << x << " "; cerr << endl;} }

constexpr int INF = 1e9+5;
constexpr long long INFLL = 1LL<<60;

class UnionFind {
public:
  // 親の番号を格納する。親だった場合は-(その集合のサイズ)
  vector<int> parent;

  // 作るときは parent の値を全て -1 にする
  // こうすると全てバラバラになる
  UnionFind(const int& n = 0) : parent(n, -1) {}

  // x がどのグループに属しているか調べる
  int find(const int& x) {
    if (parent[x] < 0) return x;
    return parent[x] = find(parent[x]);
  }

  // x と y をくっつける  O(log n)
  bool unite(int x, int y) {
    // x と y を直接くっつけるのではなく，root(x) に root(y) をくっつける
    x = find(x);
    y = find(y);
    if (x == y) return false;  // すでにくっついているからくっつけない

    // 大きい方(x) に小さい方(y)をくっつけたい 大小が逆だったらひっくり返す
    if (size(x) < size(y)) swap(x, y);

    parent[x] += parent[y];  // x のサイズを更新する
    parent[y] = x;  // y の親を x に変更する
    return true;
  }

  // 自分のいるグループの頂点数を調べる
  int size(const int& x) { return -parent[find(x)]; }

  // x と y が同じグループに属しているか否か
  bool same(const int& x, const int& y) { return find(x) == find(y); }
};

using P = pair<int, int>;

// a 番目の edge が橋なら true を返す
bool check(const vector<P>& edges, const int& n, int a) {
  UnionFind uf(n);
  for (int i = 0; i < edges.size(); i++) {
    if (i == a) continue;
    uf.unite(edges[i].first, edges[i].second);
  }
  return uf.size(0) != n;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, m;
  cin >> n >> m;

  vector<P> edges;
  rep(i, m) {
    int a, b; cin >> a >> b; a--; b--;
    edges.push_back({a, b});
  }

  int ans = 0;
  rep(i, m) if (check(edges, n, i)) ans++;

  cout << ans << endl;

  return 0;
}