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.ArrayDeque;
import java.util.Arrays;
import java.util.Deque;
import java.util.Scanner;

public class Main {
	int count = 1;
	int ans = 0;
	void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		boolean[] flag = new boolean[n+1];
		int[] a = new int[n];
		int[] b = new int[n];
		int[][] V = new int[n+1][n+1];
		int m = sc.nextInt();
		for(int i = 0; i < m; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
			V[a[i]][b[i]] = V[b[i]][a[i]] = 1;
		}
		Deque<Integer> q = new ArrayDeque<>();
		for(int i = 0; i < m; i++) {
			V[a[i]][b[i]] = V[b[i]][a[i]] = 0;
			q.push(1);
			flag[a[0]] = true;
			while(!q.isEmpty()) {
				int tmp = q.pop();
				for(int j = 1; j <= n; j++) {
					if(V[tmp][j] == 1 && !flag[j]) {
						q.push(j);
						flag[j] = true;
						count++;
					}
				}
			}
			if(count != n) {
				ans++;
			}
			count = 1;
			V[a[i]][b[i]] = V[b[i]][a[i]] = 1;
			Arrays.fill(flag, false);;
		}
		System.out.println(ans);
		sc.close();
	}

	public static void main(String[] args) {
		new Main().run();
	}
}