#include<bits/stdc++.h>
#define N 100005
using namespace std;
typedef pair<int,int> P;
int n,m,s,t,A[N],used[N],depth[N];
vector<int> G[N];
vector<P> ans;

void dfs(int x,int d,int prev){
  depth[x]=d;
  for(int i=0;i<G[x].size();i++){
    int nx=G[x][i];
    if(nx==prev)continue;
    if(!used[nx]){
      used[nx]=true;
      dfs(nx,d+1,x);
      if(!A[nx])
	ans.push_back(P(min(x,nx),max(x,nx)));
      A[x]+=A[nx];
    }
    else if(d>depth[nx])A[nx]--,A[x]++;
  }
}

int main(){
  cin>>n>>m;
  for(int i=0;i<m;i++){
    cin>>s>>t;
    G[s].push_back(t);
    G[t].push_back(s);
  }
  used[0]=1;
  dfs(0,0,0);
  sort(ans.begin(),ans.end());
  for(int i=0;i<ans.size();i++)
    cout<<ans[i].first<<' '<<ans[i].second<<endl;
  return 0;
}
### Prompt

Generate 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>
#define N 100005
using namespace std;
typedef pair<int,int> P;
int n,m,s,t,A[N],used[N],depth[N];
vector<int> G[N];
vector<P> ans;

void dfs(int x,int d,int prev){
  depth[x]=d;
  for(int i=0;i<G[x].size();i++){
    int nx=G[x][i];
    if(nx==prev)continue;
    if(!used[nx]){
      used[nx]=true;
      dfs(nx,d+1,x);
      if(!A[nx])
	ans.push_back(P(min(x,nx),max(x,nx)));
      A[x]+=A[nx];
    }
    else if(d>depth[nx])A[nx]--,A[x]++;
  }
}

int main(){
  cin>>n>>m;
  for(int i=0;i<m;i++){
    cin>>s>>t;
    G[s].push_back(t);
    G[t].push_back(s);
  }
  used[0]=1;
  dfs(0,0,0);
  sort(ans.begin(),ans.end());
  for(int i=0;i<ans.size();i++)
    cout<<ans[i].first<<' '<<ans[i].second<<endl;
  return 0;
}
```