#include<bits/stdc++.h>
using namespace std;
vector<int> v[100000];
vector<pair<int,int> > ans;
int S[100000],D[100000];

void dfs(int p,int d,int pr){
  D[p]=d;
  for(int i=0;i<v[p].size();i++){
    int np=v[p][i];
    if(pr==np)continue;
    if(!D[np]){
      dfs(np,d+1,p);
      if(!S[np])ans.push_back(make_pair(min(p,np),max(p,np)));
      S[p]+=S[np];
    }
    else if(D[p]>D[np])S[np]--,S[p]++;
  }
}

int main(){
  int V,E;
  cin>>V>>E;
  for(int s,t,i=0;i<E;i++)
    cin>>s>>t,v[s].push_back(t),v[t].push_back(s);
  dfs(0,1,-1);
  sort(ans.begin(),ans.end());
  for(int i=0;i<ans.size();i++)cout<<ans[i].first<<" "<<ans[i].second<<endl;
  return 0;
}
### Prompt

Your task is to create a cpp solution to the following problem:

Find bridges of an undirected graph G(V, E).

A bridge (also known as a cut-edge) is an edge whose deletion increase the number of connected components.

Constraints

* 1 ≤ |V| ≤ 100,000
* 0 ≤ |E| ≤ 100,000
* The graph is connected
* There are no parallel edges
* There are no self-loops

Input


|V| |E|
s0 t0
s1 t1
:
s|E|-1 t|E|-1


, where |V| is the number of nodes and |E| is the number of edges in the graph. The graph nodes are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target nodes of i-th edge (undirected).

Output

A list of bridges of the graph ordered by name. For each bridge, names of its end-ponints, source and target (source < target), should be printed separated by a space. The sources should be printed ascending order, then the target should also be printed ascending order for the same source.

Examples

Input

4 4
0 1
0 2
1 2
2 3


Output

2 3


Input

5 4
0 1
1 2
2 3
3 4


Output

0 1
1 2
2 3
3 4

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
vector<int> v[100000];
vector<pair<int,int> > ans;
int S[100000],D[100000];

void dfs(int p,int d,int pr){
  D[p]=d;
  for(int i=0;i<v[p].size();i++){
    int np=v[p][i];
    if(pr==np)continue;
    if(!D[np]){
      dfs(np,d+1,p);
      if(!S[np])ans.push_back(make_pair(min(p,np),max(p,np)));
      S[p]+=S[np];
    }
    else if(D[p]>D[np])S[np]--,S[p]++;
  }
}

int main(){
  int V,E;
  cin>>V>>E;
  for(int s,t,i=0;i<E;i++)
    cin>>s>>t,v[s].push_back(t),v[t].push_back(s);
  dfs(0,1,-1);
  sort(ans.begin(),ans.end());
  for(int i=0;i<ans.size();i++)cout<<ans[i].first<<" "<<ans[i].second<<endl;
  return 0;
}
```