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"stdio.h"
#include"iostream"
#include"vector"

#define V_MAX 50
using namespace std;

long long w[10000], N, W,M,a[51],b[51], rec,ans;
bool use_v[V_MAX], use_e[V_MAX+1][V_MAX];
long long max(long long x, long long y){
	if (x > y)return x;
	return y;

}

int min(int x, int y){
	if (x > y)return y;
	return x;
}

void dfs(int v){
	use_v[v] = true;
	for (int i = 0; i < N; i++){
		if (use_e[v][i]){
			if (!use_v[i]){
				dfs(i);
			}
		}
	}
	return;
}


int main(){
	cin >> N >> M;
	for (int i = 0; i <= N; i++){
		use_v[i]=false;
	}
	for (int i = 0; i < M; i++){
		cin >> a[i] >> b[i];
		a[i]--;
		b[i]--;
		use_e[a[i]][b[i]]=use_e[b[i]][a[i]]=true;
	}
	ans = 0;
	for (int k = 0; k < M; k++){
		use_e[a[k]][b[k]] = use_e[b[k]][a[k]] = false;
		for (int j = 0; j < N; j++) use_v[j] = false;
		dfs(0);
		for (int j = 0; j < N; j++){
			if (!use_v[j]){
				ans++;
				break;
			}
		}
		use_e[a[k]][b[k]] = use_e[b[k]][a[k]] = true;
	}
	cout << ans << endl;
	return 0;
}