#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>

using namespace std;

vector<pair<int,int>> bridges;
vector<int> graph[100000];
int ord[100000],low[100000];
bool visited[100000];

int visit_time = 0;

void dfs(int v,int p){
  visited[v] = true;
  ord[v] = visit_time;
  low[v] = ord[v];
  visit_time++;
  for(int i=0;i<graph[v].size();++i){
    if(!visited[graph[v][i]]){
      dfs(graph[v][i],v);
      low[v] = min(low[v],low[graph[v][i]]);
      if(ord[v] < low[graph[v][i]]){
        int s = v, t = graph[v][i];
        if(s > t) swap(s,t);
        bridges.emplace_back(s,t);
      }
    }else if(graph[v][i]!=p){
      low[v] = min(low[v],ord[graph[v][i]]);
    }
  }

}

int main(){
  int N , M ;
  cin >> N >> M;
  for(int i=0;i<M;++i){
    int s,t;
    cin >> s >> t;
    graph[s].push_back(t);   
    graph[t].push_back(s);
  }
  
  for(int i=0;i<N;++i){
    if(!visited[i]) dfs(i,-1);
  }
  
  sort(bridges.begin(),bridges.end());
  for(const auto &b:bridges){
    cout << b.first << " " << b.second << endl;
  }
  return 0;
}
### Prompt

Please create a solution in Cpp 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 <iostream>
#include <algorithm>
#include <vector>
#include <utility>

using namespace std;

vector<pair<int,int>> bridges;
vector<int> graph[100000];
int ord[100000],low[100000];
bool visited[100000];

int visit_time = 0;

void dfs(int v,int p){
  visited[v] = true;
  ord[v] = visit_time;
  low[v] = ord[v];
  visit_time++;
  for(int i=0;i<graph[v].size();++i){
    if(!visited[graph[v][i]]){
      dfs(graph[v][i],v);
      low[v] = min(low[v],low[graph[v][i]]);
      if(ord[v] < low[graph[v][i]]){
        int s = v, t = graph[v][i];
        if(s > t) swap(s,t);
        bridges.emplace_back(s,t);
      }
    }else if(graph[v][i]!=p){
      low[v] = min(low[v],ord[graph[v][i]]);
    }
  }

}

int main(){
  int N , M ;
  cin >> N >> M;
  for(int i=0;i<M;++i){
    int s,t;
    cin >> s >> t;
    graph[s].push_back(t);   
    graph[t].push_back(s);
  }
  
  for(int i=0;i<N;++i){
    if(!visited[i]) dfs(i,-1);
  }
  
  sort(bridges.begin(),bridges.end());
  for(const auto &b:bridges){
    cout << b.first << " " << b.second << endl;
  }
  return 0;
}
```