#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <utility>
using namespace std;

vector<pair<int, int>> bridge(const vector<vector<int>> &G) {
  const int N = G.size();
  vector<bool> vis(N, false);
  vector<int> parent(N);
  vector<int> indeg(N, 0);
  vector<int> imos(N, 0);
  stack<int> S;
  S.push(0);
  parent[0] = -1;
  while(!S.empty()) {
    int v = S.top(); S.pop();
    if(vis[v]) continue;
    vis[v] = true;
    for(int u : G[v]) {
      if(u == parent[v]) continue;
      if(vis[u]) {
        imos[v]++;
        imos[u]--;
        indeg[u]--;
      }
      else {
        parent[u] = v;
        indeg[v]++;
        S.push(u);
      }
    }
  }
  queue<int> Q;
  for(int i = 0; i < N; ++i) {
    if(indeg[i] == 0) Q.push(i);
  }
  vector<pair<int, int>> ret;
  while(!Q.empty()) {
    int v = Q.front(); Q.pop();
    if(parent[v] == -1) break;
    int u = parent[v];
    if(imos[v] == 0) ret.emplace_back(v, u);
    imos[u] += imos[v];
    indeg[u]--;
    if(indeg[u] == 0) Q.push(u);
  }
  return ret;
}

int main() {
  ios::sync_with_stdio(false);
  int V, E; cin >> V >> E;
  vector<vector<int>> G(V);
  for(int i = 0; i < E; ++i) {
    int s, t; cin >> s >> t;
    G[s].push_back(t);
    G[t].push_back(s);
  }
  auto &&ret = bridge(G);
  for(auto &&item : ret) {
    if(item.first > item.second) swap(item.first, item.second);
  }
  sort(ret.begin(), ret.end());
  for(auto &&item : ret) {
    cout << item.first << ' ' << item.second << endl;
  }
}
### Prompt

Please formulate 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 <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <utility>
using namespace std;

vector<pair<int, int>> bridge(const vector<vector<int>> &G) {
  const int N = G.size();
  vector<bool> vis(N, false);
  vector<int> parent(N);
  vector<int> indeg(N, 0);
  vector<int> imos(N, 0);
  stack<int> S;
  S.push(0);
  parent[0] = -1;
  while(!S.empty()) {
    int v = S.top(); S.pop();
    if(vis[v]) continue;
    vis[v] = true;
    for(int u : G[v]) {
      if(u == parent[v]) continue;
      if(vis[u]) {
        imos[v]++;
        imos[u]--;
        indeg[u]--;
      }
      else {
        parent[u] = v;
        indeg[v]++;
        S.push(u);
      }
    }
  }
  queue<int> Q;
  for(int i = 0; i < N; ++i) {
    if(indeg[i] == 0) Q.push(i);
  }
  vector<pair<int, int>> ret;
  while(!Q.empty()) {
    int v = Q.front(); Q.pop();
    if(parent[v] == -1) break;
    int u = parent[v];
    if(imos[v] == 0) ret.emplace_back(v, u);
    imos[u] += imos[v];
    indeg[u]--;
    if(indeg[u] == 0) Q.push(u);
  }
  return ret;
}

int main() {
  ios::sync_with_stdio(false);
  int V, E; cin >> V >> E;
  vector<vector<int>> G(V);
  for(int i = 0; i < E; ++i) {
    int s, t; cin >> s >> t;
    G[s].push_back(t);
    G[t].push_back(s);
  }
  auto &&ret = bridge(G);
  for(auto &&item : ret) {
    if(item.first > item.second) swap(item.first, item.second);
  }
  sort(ret.begin(), ret.end());
  for(auto &&item : ret) {
    cout << item.first << ' ' << item.second << endl;
  }
}
```