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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

/**
 * 解説通りに作成したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt(), m = scanner.nextInt();
			int[] a = new int[m], b = new int[m];
			boolean[][] graph = new boolean[n][n];
			IntStream.range(0, n).forEach(i -> Arrays.fill(graph[i], false));
			IntStream.range(0, m).forEach(i -> {
				a[i] = scanner.nextInt() - 1;
				b[i] = scanner.nextInt() - 1;
				graph[a[i]][b[i]] = graph[b[i]][a[i]] = true;
			});
			boolean[] visited = new boolean[n];
			System.out.println(IntStream.range(0, m).filter(i -> {
				Arrays.fill(visited, false);
				graph[a[i]][b[i]] = graph[b[i]][a[i]] = false;
				dnf(graph, visited, 0);
				graph[a[i]][b[i]] = graph[b[i]][a[i]] = true;
				return IntStream.range(0, n).filter(j -> !visited[j]).count() > 0;
			}).count());
		}
	}

	/**
	 * グラフの遍歴処理
	 * 
	 * @param graph   グラフを表す配列
	 * @param visited 各ノードが処理されたかどうかを表す配列
	 * @param node    現在処理しているノード
	 */
	private static void dnf(final boolean[][] graph, boolean[] visited, int node) {
		visited[node] = true;
		IntStream.range(0, visited.length).filter(i -> graph[node][i]).forEach(i -> {
			if (!visited[i]) {
				dnf(graph, visited, i);
			}
		});
	}
}