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:
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>

using namespace std;

typedef pair<int,int> P;

const int MAX=100000;

int n;
vector<int> G[MAX];
int order[MAX]={};
bool inS[MAX]={};
stack<int> roots,S;
int k;
vector<P> brg;

void visit(int v,int p)
{
	order[v]=++k;
	inS[v]=true;
	S.push(v);
	roots.push(v);
	for(int i=0;i<G[v].size();i++)
	{
		int to=G[v][i];
		if(order[to]==0)
		{
			visit(to,v);
		}
		else if(to!=p&&inS[to])
		{
			while(order[roots.top()]>order[to])roots.pop();
		}
	}
	if(v==roots.top())
	{
		if(p!=-1)brg.push_back(P(min(p,v),max(p,v)));
		while(1)
		{
			int node=S.top();S.pop();
			inS[node]=false;
			if(node==v)break;
		}
		roots.pop();
	}
	return;
}

void bridge()
{
	k=0;
	for(int i=0;i<n;i++)
	{
		if(order[i]==0)
		{
			visit(i,-1);
		}
	}
	return;
}

int main()
{
	cin>>n;
	int e;
	cin>>e;
	int u,v;
	for(int i=0;i<e;i++)
	{
		cin>>u>>v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	bridge();
	sort(brg.begin(),brg.end());
	for(int i=0;i<brg.size();i++)
	{
		cout<<brg[i].first<<" "<<brg[i].second<<endl;
	}
	return 0;
}