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

int V, E;

void dfs(vector<vector<int>> &G, int p, int u, int d, vector<int> &depth, vector<int> &lowlink, vector<bool> &visited){
    visited[u] = true;
    depth[u] = lowlink[u] = d;
    
    for(auto v: G[u]) if(v != p){
        if(!visited[v]){
            dfs(G, u, v, d+1, depth, lowlink, visited);
            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);
    }
    for(int i=0; i<V; i++)
        sort(G[i].begin(), G[i].end());
    
    vector<int> depth(V, -1), lowlink(V, -1);
    vector<bool> visited(V, false);
    dfs(G, -1, 0, 0, depth, lowlink, visited);
    
    for(int s=0; s<V; s++){
        for(auto t: G[s]) if(s<t && (depth[s]<lowlink[t]||depth[t]<lowlink[s])){
            cout << s << " " << t << endl;
        }
    }
    
    return 0;
}
### Prompt

Your challenge is to write 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 <bits/stdc++.h>
using namespace std;

int V, E;

void dfs(vector<vector<int>> &G, int p, int u, int d, vector<int> &depth, vector<int> &lowlink, vector<bool> &visited){
    visited[u] = true;
    depth[u] = lowlink[u] = d;
    
    for(auto v: G[u]) if(v != p){
        if(!visited[v]){
            dfs(G, u, v, d+1, depth, lowlink, visited);
            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);
    }
    for(int i=0; i<V; i++)
        sort(G[i].begin(), G[i].end());
    
    vector<int> depth(V, -1), lowlink(V, -1);
    vector<bool> visited(V, false);
    dfs(G, -1, 0, 0, depth, lowlink, visited);
    
    for(int s=0; s<V; s++){
        for(auto t: G[s]) if(s<t && (depth[s]<lowlink[t]||depth[t]<lowlink[s])){
            cout << s << " " << t << endl;
        }
    }
    
    return 0;
}
```