#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> P;
const int MAX_V=100000;
vector<int> G[MAX_V];
int ord[MAX_V];
bool inS[MAX_V];
stack<int> roots,S;
vector<vi> cc;
vector<P> bs;
int cnt;
int N,M;
void vis(int v,int p){
	ord[v]=++cnt;
	S.push(v);
	inS[v]=1;
	roots.push(v);
	for(int u:G[v]){
		if(ord[u]==0){
			vis(u,v);
		}else if(u!=p&&inS[u]){
			while(ord[roots.top()]>ord[u]) roots.pop();
		}
	}
	if(v==roots.top()){
		bs.pb(P(p,v));
		vector<int> vc;
		while(true){
			int w=S.top();S.pop();
			inS[w]=0;
			vc.pb(w);
			if(v==w) break;
		}
		roots.pop();
		cc.pb(vc);
	}
}
void bridge(){
	rep(i,N) if(ord[i]==0){
		vis(i,-1);
		bs.pop_back();
	}
}
int main(){
	cin>>N>>M;
	rep(i,M){
		int a,b;
		cin>>a>>b;
		G[a].pb(b);
		G[b].pb(a);
	}
	bridge();
	for(P& p:bs) if(p.fs>p.sc) swap(p.fs,p.sc);
	sort(all(bs));
	for(P p:bs){
		cout<<p.fs<<" "<<p.sc<<endl;
	}
}
### Prompt

Create a solution in CPP for 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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> P;
const int MAX_V=100000;
vector<int> G[MAX_V];
int ord[MAX_V];
bool inS[MAX_V];
stack<int> roots,S;
vector<vi> cc;
vector<P> bs;
int cnt;
int N,M;
void vis(int v,int p){
	ord[v]=++cnt;
	S.push(v);
	inS[v]=1;
	roots.push(v);
	for(int u:G[v]){
		if(ord[u]==0){
			vis(u,v);
		}else if(u!=p&&inS[u]){
			while(ord[roots.top()]>ord[u]) roots.pop();
		}
	}
	if(v==roots.top()){
		bs.pb(P(p,v));
		vector<int> vc;
		while(true){
			int w=S.top();S.pop();
			inS[w]=0;
			vc.pb(w);
			if(v==w) break;
		}
		roots.pop();
		cc.pb(vc);
	}
}
void bridge(){
	rep(i,N) if(ord[i]==0){
		vis(i,-1);
		bs.pop_back();
	}
}
int main(){
	cin>>N>>M;
	rep(i,M){
		int a,b;
		cin>>a>>b;
		G[a].pb(b);
		G[b].pb(a);
	}
	bridge();
	for(P& p:bs) if(p.fs>p.sc) swap(p.fs,p.sc);
	sort(all(bs));
	for(P p:bs){
		cout<<p.fs<<" "<<p.sc<<endl;
	}
}
```