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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int N, M;
int a[64 * 64], b[64 * 64];

char edge[64][64];
char loop[64][64];

char visited[64];
int trace[64 * 64];

void dfs(int node, int depth, int parent) {
	int i;
	trace[depth] = node;
	if (visited[node]) {
		for (i = depth - 1; i >= 0; i--) {
			loop[trace[i]][trace[i + 1]] = 1;
			if (trace[i] == node) break;
		}
	} else {
		visited[node] = 1;
		for (i = 1; i <= N; i++) {
			if (i != parent && edge[node][i]) {
				dfs(i, depth + 1, node);
			}
		}
		visited[node] = 0;
	}
}

int main(void) {
	int i;
	int sumipe = 0;
	if (scanf("%d%d", &N, &M) != 2) return 1;
	for (i = 0; i < M; i++) {
		if (scanf("%d%d", &a[i], &b[i]) != 2) return 1;
		edge[a[i]][b[i]] = edge[b[i]][a[i]] = 1;
	}
	dfs(1, 0, 0);
	for (i = 0; i < M; i++) {
		if (!loop[a[i]][b[i]] && !loop[b[i]][a[i]]) sumipe++;
	}
	printf("%d\n", sumipe);
	return 0;
}