#include <iostream>
#include <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
typedef pair<double, double> P;

int dfs(int u, int prev, int *cnt, int *prenum, int *lowest, vector< vector<int> > &v, vector<P> &res){
  prenum[u] = lowest[u] = ++(*cnt);
  REP(i, v[u].size()){
    int next = v[u][i];
    if(prenum[next] == -1){
      lowest[u] = min(lowest[u], dfs(next, u, cnt, prenum, lowest, v, res));
      if(lowest[next] == prenum[next]) res.push_back(P(min(u, next), max(u, next)));
    }
    else if(prev != next)
      lowest[u] = min(lowest[u], lowest[next]);
  }
  return lowest[u];
}

vector<P> articulation_points(int V, int E, vector< vector<int> > &v){
  int prenum[V], lowest[V];
  memset(prenum, -1, sizeof(prenum));
  int tmp = 1;
  vector<P> res;
  dfs(0, -1, &tmp, prenum, lowest, v, res);
  return res;
}

int main() {
  int V, E; cin >>V >>E;
  vector< vector<int> > es(V);
  REP(i, E){
    int f, t; cin >>f >>t;
    es[f].push_back(t);
    es[t].push_back(f);
  }
  vector<P> ans = articulation_points(V, E, es);
  SORT(ans);
  REP(i, ans.size()) cout <<ans[i].first <<" " <<ans[i].second <<endl;
  return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
typedef pair<double, double> P;

int dfs(int u, int prev, int *cnt, int *prenum, int *lowest, vector< vector<int> > &v, vector<P> &res){
  prenum[u] = lowest[u] = ++(*cnt);
  REP(i, v[u].size()){
    int next = v[u][i];
    if(prenum[next] == -1){
      lowest[u] = min(lowest[u], dfs(next, u, cnt, prenum, lowest, v, res));
      if(lowest[next] == prenum[next]) res.push_back(P(min(u, next), max(u, next)));
    }
    else if(prev != next)
      lowest[u] = min(lowest[u], lowest[next]);
  }
  return lowest[u];
}

vector<P> articulation_points(int V, int E, vector< vector<int> > &v){
  int prenum[V], lowest[V];
  memset(prenum, -1, sizeof(prenum));
  int tmp = 1;
  vector<P> res;
  dfs(0, -1, &tmp, prenum, lowest, v, res);
  return res;
}

int main() {
  int V, E; cin >>V >>E;
  vector< vector<int> > es(V);
  REP(i, E){
    int f, t; cin >>f >>t;
    es[f].push_back(t);
    es[t].push_back(f);
  }
  vector<P> ans = articulation_points(V, E, es);
  SORT(ans);
  REP(i, ans.size()) cout <<ans[i].first <<" " <<ans[i].second <<endl;
  return 0;
}
```