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>
#include <stdio.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()

typedef long long ll;
typedef pair<ll, int> P;

int n,m,sum;
int r[50];
vector<int> v[50];
bool used[50];

void cnt(int idx) {
	used[idx]=true;
	sum++;
	int nxt,ccnt=0;
	REP(i,v[idx].size()) {
		if(used[v[idx][i]]) continue;
		ccnt++;
		nxt=v[idx][i];
	}
	if(ccnt==1) {
		cnt(nxt);
	}
	else sum--,used[idx]=false;
	return;
}

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

	REP(i,50) used[i]=false;

	cin>>n>>m;
	REP(i,m) {
		int a,b; cin>>a>>b; a--;b--;
		r[a]++; r[b]++;
		v[a].push_back(b); v[b].push_back(a);
	}
	REP(i,n) {
		if(r[i]==1) {
			cnt(i);
		}
	}
	cout<<sum<<endl;

	return 0;
}