#include<bits/stdc++.h>

using namespace std;

const int SIZE = 100013;

typedef pair<int, int> pii;

bool visited[SIZE];
int disc[SIZE], low[SIZE], parent[SIZE];

vector<int> G[SIZE];


void bridges(int u, priority_queue<pii, vector<pii>, greater<pii> > *bridge){
  static int time = 0;
  int children = 0;
  visited[u] = true;
  disc[u] = low[u] = ++time;
  for(int i = 0; i < G[u].size(); i++){
    int v = G[u][i];
    if(!visited[v]){
      children++;
      parent[v] = u;
      bridges(v, bridge);
      low[u] = min(low[u], low[v]);
      if(low[v] > disc[u]) bridge->push({min(u,v),max(u,v)});
    }
    else if(v != parent[u])
      low[u] = min(low[u], disc[v]);
  }
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  #ifdef Larra
  freopen("in","r",stdin);
  freopen("out","w",stdout);
  #endif
  int n,m;
  cin>>n>>m;
  for(int i = 0; i < m; i++){
      int u,v;
      cin>>u>>v;
      G[u].push_back(v);
      G[v].push_back(u);
  }
  priority_queue<pii, vector<pii>, greater<pii> > bridge;
  for(int i = 0; i < SIZE; i++)
    parent[i] = -1;
  bridges(0, &bridge);
  while(!bridge.empty()){
    cout<<bridge.top().first<<" "<<bridge.top().second<<"\n";
    bridge.pop();
  }
}
### 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<bits/stdc++.h>

using namespace std;

const int SIZE = 100013;

typedef pair<int, int> pii;

bool visited[SIZE];
int disc[SIZE], low[SIZE], parent[SIZE];

vector<int> G[SIZE];


void bridges(int u, priority_queue<pii, vector<pii>, greater<pii> > *bridge){
  static int time = 0;
  int children = 0;
  visited[u] = true;
  disc[u] = low[u] = ++time;
  for(int i = 0; i < G[u].size(); i++){
    int v = G[u][i];
    if(!visited[v]){
      children++;
      parent[v] = u;
      bridges(v, bridge);
      low[u] = min(low[u], low[v]);
      if(low[v] > disc[u]) bridge->push({min(u,v),max(u,v)});
    }
    else if(v != parent[u])
      low[u] = min(low[u], disc[v]);
  }
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  #ifdef Larra
  freopen("in","r",stdin);
  freopen("out","w",stdout);
  #endif
  int n,m;
  cin>>n>>m;
  for(int i = 0; i < m; i++){
      int u,v;
      cin>>u>>v;
      G[u].push_back(v);
      G[v].push_back(u);
  }
  priority_queue<pii, vector<pii>, greater<pii> > bridge;
  for(int i = 0; i < SIZE; i++)
    parent[i] = -1;
  bridges(0, &bridge);
  while(!bridge.empty()){
    cout<<bridge.top().first<<" "<<bridge.top().second<<"\n";
    bridge.pop();
  }
}
```