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> // cin, cout, cerr, clog
#include <algorithm> // minmax, sort, swap
#include <numeric> // iota, gcd, lcm 
#include <cstdio> // printf, scanf
#include <climits> // INT_MIN, LLONG_MIN
#include <cmath> // long, trig, pow
#include <vector> // vectors
#include <string> // string
#include <queue> // queue
#include <stack> // stack
#include <map> // key-value pairs sorted by keys
#include <unordered_map> // hashed by keys
#include <iomanip> //cout<<setprecision(n)
#define rep(i,n) for(int i = 0; i < n; i++)

typedef long long ll; // at least int64 > 9*10^18
typedef unsigned long long ull; // at least uint64

struct UnionFind
{
  int n;
  std::vector<int> par,size;
  void reset() {
    std::iota(par.begin(),par.end(),0);
  }
  UnionFind(int sz) : n(sz), par(sz,0), size(sz,0) {
    reset();
  }
  int find(int x) {
    if(par[x]==x) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x=find(x);y=find(y);
    if(x==y) return;
    if(size[x]<size[y]) std::swap(x,y);
    size[x]+=size[y];
    par[y]=x;
  }
};

int main() {
  int n,m;
  std::cin >> n >> m;
  int a[m];
  int b[m];
  rep(i,m) {
    std::cin >> a[i] >> b[i];
    a[i]--;
    b[i]--;
  }
  int count = 0;
  UnionFind uf(n);
  rep(i,m) {
    uf.reset();
    rep(j,m) {
      if(i!=j) uf.unite(a[j],b[j]);
    }
    bool connected = true;
    int tmp = uf.find(0);
    rep(j,n) {
      if(uf.find(j)!=tmp) connected = false;
    }
    if(connected == false) count ++;
  }
  std::cout << count << std::endl;
  return 0;
}