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

Solution (in CPP):

Solution:
//In His Name
#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >e;
vector<int>adj[100100];
int n,m,minlvl[100100],lvl[100100];
bool mark[100100];
void dfs(int x,int par){
	mark[x]=true;
	minlvl[x]=lvl[x];
	for(int i=0;i<adj[x].size();i++){
		int v=adj[x][i];
		if(!mark[v]){
			lvl[v]=lvl[x]+1;
			dfs(v,x);
			minlvl[x]=min(minlvl[x],minlvl[v]);
		}
		else if(v!=par) minlvl[x]=min(minlvl[x],minlvl[v]);
	}
}
int main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int x,y;
		cin>>x>>y;
		if(x>y) swap(x,y);
		adj[x].push_back(y);
		adj[y].push_back(x);
		e.push_back(make_pair(x,y));
	}
	for(int i=0;i<n;i++)
		if(!mark[i]) dfs(i,i);
	sort(e.begin(),e.end());
	for(int i=0;i<m;i++){
		int x=e[i].first,y=e[i].second;
		if(lvl[x]<lvl[y]) swap(x,y);
		if(minlvl[x]>lvl[y]) cout<<e[i].first<<" "<<e[i].second<<endl;
	}
	return -0;
}