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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;

public class Main {

	static int n, m;
	static int[] a, b;
	static boolean[] vis;
	static boolean[][] g;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		m = sc.nextInt();
		a = new int[n];
		b = new int[n];
		for(int i = 0 ; i < m ; i++) {
			a[i] = sc.nextInt() - 1;
			b[i] = sc.nextInt() - 1;
		}
		vis = new boolean[n];
		g = new boolean[n][n];
		for(int i = 0 ; i < m ; i++) {
			g[a[i]][b[i]] = g[b[i]][a[i]] = true;
		}
		int ans = 0;
		for(int i = 0 ; i < m ; i++) {
			for(int j = 0 ; j < n ; j++) vis[j] = false;
			g[a[i]][b[i]] = g[b[i]][a[i]] = false;
			dfs(0);
			boolean bridge = false;
			for(int j = 0 ; j < n ; j++) {
				if(!vis[j]) bridge = true;
			}
			if(bridge) ans++;
			g[a[i]][b[i]] = g[b[i]][a[i]] = true;
		}
		System.out.println(ans);
	}

	static void dfs(int x) {
		vis[x] = true;
		for(int i = 0 ; i < n ; i++) {
			if(!vis[i] && g[x][i]) dfs(i);
		}
	}

}