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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

/*ソートの時に使う
int sort(const void *a, const void *b){
    return *(int*)a-*(int*)b ;
}
    qsort (data,sizeof(data),sizeof(int),sort); 
*/

int main(void){
    int n,m;
    scanf("%d%d",&n,&m);
    int a[m],b[m],p[n+1],ans=0;
    for (int i=0; i<m; i++){
        scanf("%d%d",&a[i],&b[i]);
    }
    
    for (int i=0; i<n+1; i++){
        p[i]=0;
        for(int j=0; j<n; j++){
            if (a[j]==i){
                p[i]++;
            }
            if (b[j]==i){
                p[i]++;
            }
        }
    }
    int count=1;
    while(count!=0){
        count=0;
        for (int i=0; i<m; i++){
            if (p[a[i]]==1 || p[b[i]]==1){
                ans++;
                p[a[i]]--;
                p[b[i]]--;
                count++;

            }
        }
    }
    printf ("%d\n",ans);
	return 0 ;
}