#include<bits/stdc++.h>
using namespace std;
const int M=1e5+10;
int n,m,H[M];
vector<int>g[M];
bool cv[M],mark[M];
vector<pair<int,int> >ans;
int dfs(int v,int par,int h)
{
  int up=h;
  mark[v]=true;
  H[v]=h;
  for(auto u:g[v])
    if(!mark[u])
      {
	int d=dfs(u,v,h+1);
	up=min(up,d);
	if(d>h)ans.push_back({min(u,v),max(u,v)});
      }
    else if(u!=par)
      up=min(up,H[u]);
  return up;
}
int main()
{
  cin>>n>>m;
  for(int i=0;i<m;i++)
    {
      int u,v;
      cin>>u>>v;
      u++,v++;
      g[u].push_back(v);
      g[v].push_back(u);
    }
  dfs(1,1,0);
  sort(ans.begin(),ans.end());
  for(auto p:ans)
    cout<<p.first-1<<" "<<p.second-1<<endl;
  return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int M=1e5+10;
int n,m,H[M];
vector<int>g[M];
bool cv[M],mark[M];
vector<pair<int,int> >ans;
int dfs(int v,int par,int h)
{
  int up=h;
  mark[v]=true;
  H[v]=h;
  for(auto u:g[v])
    if(!mark[u])
      {
	int d=dfs(u,v,h+1);
	up=min(up,d);
	if(d>h)ans.push_back({min(u,v),max(u,v)});
      }
    else if(u!=par)
      up=min(up,H[u]);
  return up;
}
int main()
{
  cin>>n>>m;
  for(int i=0;i<m;i++)
    {
      int u,v;
      cin>>u>>v;
      u++,v++;
      g[u].push_back(v);
      g[v].push_back(u);
    }
  dfs(1,1,0);
  sort(ans.begin(),ans.end());
  for(auto p:ans)
    cout<<p.first-1<<" "<<p.second-1<<endl;
  return 0;
}
```