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;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

//#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }

struct UnionFind {
	vector<int> parent;
	int size;
	UnionFind(int n) :parent(n, -1), size(n) {}
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x == y)return false;
		if (sizeOf(x) < sizeOf(y))swap(x, y);
		parent[x] += parent[y]; parent[y] = x; size--;
		return true;
	}
	bool same(int x, int y) { return root(x) == root(y); }
	int root(int x) { return parent[x] < 0 ? x : parent[x] = root(parent[x]); }
	int sizeOf(int x) { return -parent[root(x)]; }
};



signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N; cin >> N;
	int M; cin >> M;
	vector<int> a(M), b(M); rep(i, 0, M) {
		cin >> a[i] >> b[i];
		a[i]--, b[i]--;
	}
	int ans = 0;
	rep(i, 0, M) {
		UnionFind uf(N);
		rep(j, 0, M) {
			if (i == j)continue;
			uf.unite(a[j], b[j]);
		}
		if (uf.size != 1)
			ans++;
	}
	cout << ans << endl;
	return 0;
}