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 <bits/stdc++.h>
using namespace std;

vector<int> X;
vector<vector<int>> Y;
int ans;

void dfs(int x) {
  if (X.at(x) != 1) return;
  int y = 1;
  while(!Y.at(x).at(y)) y++;
  X.at(x)--, X.at(y)--;
  Y.at(x).at(y) = Y.at(y).at(x) = 0;
  ans++;
  dfs(y);
}

int main() {
  int N, M;
  cin >> N >> M;
  X.resize(N + 1);
  Y.resize(N + 1, vector<int>(N + 1));
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    X.at(a)++, X.at(b)++;
    Y.at(a).at(b)++, Y.at(b).at(a)++;
  }

  for (int i = 1; i <= N; i++) dfs(i);

  cout << ans << "\n";
}