#include<bits/stdc++.h>
using namespace std;
const int MAXN = 100010;
vector<int> grf[MAXN];
int visited[MAXN],tin[MAXN],low[MAXN];
int timer = 0, n,m;
set<int> arp;
set<pair<int,int>> brgs;

void dfs1(int u,int p=-1){
    visited[u]=true;
    tin[u]=low[u]= timer++;
    int childs=0;
    for(auto x: grf[u]){
        if(x==p){continue;}
        if( visited[x]){
            // ancestor
            low[u]=min(low[u],tin[x]);
        }
        else{
            childs++;
            dfs1(x,u);
            low[u] = min(low[u],low[x]);
            if(p!= -1 and low[x] >= tin[u]){
                arp.insert(u);
            }
        }
    }

    if( p==-1 and childs>1){
        arp.insert(u);
    }
}

void dfs(int u,int p=-1){
    visited[u]=true;
    tin[u]=low[u]=timer++;
    for(auto x:grf[u]){
        if(x==p){continue;}
        if(visited[x]){
            // ancestor
            low[u] = min(low[u], tin[x]);
        }
        else{
            dfs(x,u);
            low[u] = min(low[u],low[x]);
            if(low[x] > tin[u]){
                brgs.insert({min(u,x),max(u,x)});
            }
        }
    }
}


int main() {
    // freopen("./input.txt","r",stdin);
    // freopen("./output.txt","w",stdout);
    cin>>n>>m;
    for(int i=0; i<m; i++){
        int x,y; cin>>x>>y;
        grf[x].push_back(y);
        grf[y].push_back(x);
    }
    for(int i=0; i<n; i++){
        if(!visited[i]){
            dfs(i);
        }
    }
    for(auto p: brgs){
        cout<<p.first<<" "<<p.second<<"\n";
    }
}

### Prompt

Generate 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;
const int MAXN = 100010;
vector<int> grf[MAXN];
int visited[MAXN],tin[MAXN],low[MAXN];
int timer = 0, n,m;
set<int> arp;
set<pair<int,int>> brgs;

void dfs1(int u,int p=-1){
    visited[u]=true;
    tin[u]=low[u]= timer++;
    int childs=0;
    for(auto x: grf[u]){
        if(x==p){continue;}
        if( visited[x]){
            // ancestor
            low[u]=min(low[u],tin[x]);
        }
        else{
            childs++;
            dfs1(x,u);
            low[u] = min(low[u],low[x]);
            if(p!= -1 and low[x] >= tin[u]){
                arp.insert(u);
            }
        }
    }

    if( p==-1 and childs>1){
        arp.insert(u);
    }
}

void dfs(int u,int p=-1){
    visited[u]=true;
    tin[u]=low[u]=timer++;
    for(auto x:grf[u]){
        if(x==p){continue;}
        if(visited[x]){
            // ancestor
            low[u] = min(low[u], tin[x]);
        }
        else{
            dfs(x,u);
            low[u] = min(low[u],low[x]);
            if(low[x] > tin[u]){
                brgs.insert({min(u,x),max(u,x)});
            }
        }
    }
}


int main() {
    // freopen("./input.txt","r",stdin);
    // freopen("./output.txt","w",stdout);
    cin>>n>>m;
    for(int i=0; i<m; i++){
        int x,y; cin>>x>>y;
        grf[x].push_back(y);
        grf[y].push_back(x);
    }
    for(int i=0; i<n; i++){
        if(!visited[i]){
            dfs(i);
        }
    }
    for(auto p: brgs){
        cout<<p.first<<" "<<p.second<<"\n";
    }
}

```