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 MIN(int a,int b){return a<b?a:b;}
int r,ord[100010],low[100010];
void cdfs(int s,int *ta,int *to,int *nt,int *p){//printf("s%d %d\n",s,r);
  int i;
  ord[s]=low[s]=r++;
  for(i=ta[s];i+1;i=nt[i]){
    if(p[to[i]]==-1){
      p[to[i]]=s;
      cdfs(to[i],ta,to,nt,p);
      low[s]=MIN(low[s],low[to[i]]);
    }
    else if(to[i]-p[s])low[s]=MIN(low[s],ord[to[i]]);
  }
}
int brig(int v,int e,int *a,int *b,int *c,int *d){
  int ta[100010],to[100010],nt[100010],p[100010],ans=0,co=0,i,j,f[100010]={0};
  for(i=0;i<v;i++)ta[i]=p[i]=-1;
  for(p[0]=i=0;i<e;i++){
    nt[i  ]=ta[to[i+e]=a[i]];
    nt[i+e]=ta[to[i  ]=b[i]];
    ta[b[ta[a[i]]=i]]=i+e;
  }
  for(i=r=0;i<v;i++){
    if(r<v)cdfs(i,ta,to,nt,p);
  }
  for(i=1;i<v;i++){//橋を配列に格納していく
    if(ord[p[i]]<low[i]){
      c[co  ]=p[i];
      d[co++]=  i ;
    }
  }
  return co;
}
int main(){
  int n,m,i,a[110],b[110],c[110],d[110];
  scanf("%d %d",&n,&m);
  for(i=0;i<m;i++){
    scanf("%d %d",&a[i],&b[i]);
    a[i]--;
    b[i]--;
  }
  printf("%d\n",brig(n,m,a,b,c,d));
  return 0;
}