#include<bits/stdc++.h>
using namespace std;
vector<int> g[55];
int s[55],t[55];
void dfs(int used[],int k,int n){
  if(used[n])return;
  used[n]=1;
  for(int i=0;i<g[n].size();i++){
    if(n==s[k]&&g[n][i]==t[k])continue;
    if(n==t[k]&&g[n][i]==s[k])continue;
    dfs(used,k,g[n][i]);
  }
}

int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<m;i++){
    int &a=s[i],&b=t[i];
    cin>>a>>b;
    a--,b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int ans=0;
  for(int i=0;i<m;i++){
    int used[55]={};
    dfs(used,i,0);
    for(int i=0;i<n;i++){
      if(used[i]==0){
	ans++;
	break;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

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.

Examples

Input

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7


Output

4


Input

3 3
1 2
1 3
2 3


Output

0


Input

6 5
1 2
2 3
3 4
4 5
5 6


Output

5

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
vector<int> g[55];
int s[55],t[55];
void dfs(int used[],int k,int n){
  if(used[n])return;
  used[n]=1;
  for(int i=0;i<g[n].size();i++){
    if(n==s[k]&&g[n][i]==t[k])continue;
    if(n==t[k]&&g[n][i]==s[k])continue;
    dfs(used,k,g[n][i]);
  }
}

int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<m;i++){
    int &a=s[i],&b=t[i];
    cin>>a>>b;
    a--,b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int ans=0;
  for(int i=0;i<m;i++){
    int used[55]={};
    dfs(used,i,0);
    for(int i=0;i<n;i++){
      if(used[i]==0){
	ans++;
	break;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}

```