#include <bits/stdc++.h>
using namespace std;

int V, E;
vector<bool> visited;
vector<int> depth, lowlink, pars;

void dfs(vector<vector<int>> &G, int p, int u, int d){
    visited[u] = true;
    pars[u] = p;
    depth[u] = lowlink[u] = d;
    
    for(auto v: G[u]) if(v != p){
        if(!visited[v]){
            dfs(G, u, v, d+1);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }else{
            lowlink[u] = min(lowlink[u], depth[v]);
        }
    }
}

int main(){
    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);
    }
    
    depth.assign(V, -1), lowlink.assign(V, -1);
    visited.assign(V, false);
    pars.assign(V, -1);
    dfs(G, -1, 0, 0);
    
    vector<pair<int, int>> ans;
    vector<vector<int>> tree(V);
    for(int i=0; i<V; i++){
        int p = pars[i];
        if(p != -1 && depth[p]<lowlink[i]){
            if(p < i)
                ans.emplace_back(p, i);
            else
                ans.emplace_back(i, p);
        }
    }
    sort(ans.begin(), ans.end());
    
    for(auto p: ans)
        cout << p.first << " " << p.second << 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;

int V, E;
vector<bool> visited;
vector<int> depth, lowlink, pars;

void dfs(vector<vector<int>> &G, int p, int u, int d){
    visited[u] = true;
    pars[u] = p;
    depth[u] = lowlink[u] = d;
    
    for(auto v: G[u]) if(v != p){
        if(!visited[v]){
            dfs(G, u, v, d+1);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }else{
            lowlink[u] = min(lowlink[u], depth[v]);
        }
    }
}

int main(){
    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);
    }
    
    depth.assign(V, -1), lowlink.assign(V, -1);
    visited.assign(V, false);
    pars.assign(V, -1);
    dfs(G, -1, 0, 0);
    
    vector<pair<int, int>> ans;
    vector<vector<int>> tree(V);
    for(int i=0; i<V; i++){
        int p = pars[i];
        if(p != -1 && depth[p]<lowlink[i]){
            if(p < i)
                ans.emplace_back(p, i);
            else
                ans.emplace_back(i, p);
        }
    }
    sort(ans.begin(), ans.end());
    
    for(auto p: ans)
        cout << p.first << " " << p.second << endl;
    
    return 0;
}
```