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 <bits/stdc++.h>
#define N 100001
using namespace std;
typedef pair<int,int> P;
vector<int> G[N];
vector<P> ans;
int cnt[N],dep[N];

void dfs(int pos,int pre){
  for(int i=0;i<G[pos].size();i++){
    int nx=G[pos][i];
    if(nx==pre||dep[nx]>dep[pos]) continue;

    if(dep[nx]!=-1&&dep[nx]<dep[pos])cnt[nx]--,cnt[pos]++;
    else{ 
      dep[nx]=dep[pos]+1;
      dfs(nx,pos);
      if(cnt[nx]==0) ans.push_back(P(min(pos,nx),max(pos,nx)));
      cnt[pos]+=cnt[nx];
    } 
  }
}


int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0,a,b;i<m;i++){
    cin>>a>>b;
    G[a].push_back(b);
    G[b].push_back(a);
  }

  memset(dep,-1,sizeof(dep));
  dep[0]=0;
  dfs(0,-1);
  sort(ans.begin(),ans.end());
  for(int i=0;i<ans.size();i++)cout <<ans[i].first<<" "<<ans[i].second<<endl;

  return 0;
}