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:
//Have FUN

using namespace std;

#include<bits/stdc++.h>

const int maxn=1e5+10;

bool mark[maxn];
int n,m,s,t,h[maxn],up[maxn],par[maxn];
vector <int> a[maxn];

vector < pair<int,int> > ans;
vector <pair<int, int> > ed;
void input()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		cin>>s>>t;
		a[s].push_back(t);
		a[t].push_back(s);
		if(s>t)
            swap(s,t);
		ed.push_back(make_pair(s,t));
	}
}

void dfs(int x)
{
	mark[x]=true;
	up[x]=h[x];
	for(int i=0;i<a[x].size();i++)
	{
		int child=a[x][i];
		if(!mark[child])
		{
		    par[child]=x;
			h[child]=h[x]+1;
			dfs(child);
            up[x]=min(up[x],up[child]);

		}
		else
            if(child!=par[x])
                up[x]=min(up[x],h[child]);
	}
}

int main ()
{
	input();
	dfs(0);
	for(int i=0; i<m; i++)
        {
            int fi=ed[i].first, se=ed[i].second;
            if(h[fi]>h[se])
                swap(fi,se);
            if(up[se]>=h[se])
                ans.push_back(make_pair(min(fi,se), max(fi,se)));
        }
    sort(ans.begin(),ans.end());
    for(int i=0;i<ans.size();i++)
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
	return 0;
}