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>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
#include <cstdio>
#include <tuple>

#define FOR(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a,b) for(int i=a;i>b;i--)
#define vint vector<int> 
#define vvint vector<vint>
#define CST(x) cout<<fixed<<setprecision(x)//小数点以下の桁数指定
#define ct(a) cout<<a<<endl
static const double pi = 3.141592653589793;
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const int INF = 1e9+7;

class unionfind {
public:
	vector<int> par;
	vector<int> rnk;

	unionfind(int N) :par(N),rnk(N) {//コンストラクタ
		for (int i = 0; i < N; i++) {
			par[i] = i;
			rnk[i] = 0;
		}
	}

	int root(int x) {//データｘが属する木の根を再帰で得る:root(x)={ｘの木の根}
		if (par[x] == x)return x;
		return par[x] = root(par[x]);

	}

	void unite(int x, int y) {
		int rx = root(x);
		int ry = root(y);
		if (rx == ry)return;//ｘとｙの根が同じ（＝同じ木にある）時はそのまま
		if (rnk[rx] < rnk[ry]) {
			par[rx] = ry;//xとyの根が同じでない（＝同じ木にない）時：ｘの根をｒｘをｙの根につける
		}
		else if (rnk[rx] > rnk[ry]) {
			par[ry] = rx;
		}
		else {
			par[rx] = ry;
			rnk[ry]++;
		}
	}

	bool same(int x, int y) {
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}

};

bool edge[51][51];
pair<int, int>P[51];

int main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(false);
	
	int N, M;
	int ans = 0;
	cin >> N >> M;

	FOR(i, 0, M) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		P[i] = make_pair(a, b);
	}
	FOR(i, 0, M) {
		unionfind U(N);

		FOR(j, 0, M) {
			if(i!=j){
				U.unite(P[j].first, P[j].second);
			}
		}
		if (U.root(P[i].first) != U.root(P[i].second)) {
			ans++;
		}
	}




	cout << ans << endl;


	return 0;
}