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>
using namespace std;
typedef pair<int,int> P;
#define MAX 100005

int V,E;
vector<int> G[MAX];

int depth[MAX];
int cnt[MAX];
int visited[MAX];

vector<P> ans;

void dfs(int pos,int prev){
  visited[pos]=true;
  for(int i=0;i<(int)G[pos].size();i++){
    int to=G[pos][i];
    if(to==prev)continue;
    if(!visited[to]){
      depth[to]=depth[pos]+1;
      dfs(to,pos);
      cnt[pos]+=cnt[to];
      if(cnt[to]==0)ans.push_back( P( min(pos,to) , max(pos,to) ) );
    }else if(depth[to]<depth[pos]){
      cnt[pos]++;
      cnt[to]--;
    }
  }
}

int main(){
  scanf("%d %d",&V,&E);
  for(int i=0;i<E;i++){
    int a,b;
    scanf("%d %d",&a,&b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(0,-1);
  sort(ans.begin(),ans.end());
  
  for(int i=0;i<(int)ans.size();i++)
    printf("%d %d\n",ans[i].first,ans[i].second);
  return 0;
}