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<vector>
#include<string>
#include<cstring>
#include<iomanip>
#include<algorithm>
#include<list>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
const int MAX=100005;
const int inf=1e9;
vi G[MAX];
int N;
bool visited[MAX];
typedef pair<int,int> P;
int prenum[MAX],parent[MAX],lowest[MAX],timer;
vector<P> bridge;
void dfs(int current,int prev)
{
  prenum[current]=lowest[current]=timer;
  timer++;

  visited[current]=true;
  int next;
  for(int i=0;i<G[current].size();i++)
  {
    next=G[current][i];
    if(!visited[next])
    {
      parent[next]=current;
      dfs(next,current);
      lowest[current]=min(lowest[current],lowest[next]);
      if(prenum[current]<lowest[next]) bridge.push_back(P(min(current,next),max(current,next)));
    }
    else if(next!=prev)
    {
      lowest[current]=min(lowest[current],prenum[next]);
    }
  }
}

void bridge_points()
{
  timer=1;
  dfs(0,-1);
  sort(bridge.begin(),bridge.end());
  for(int i=0;i<bridge.size();i++)
  {
    cout << bridge[i].first << ' ' << bridge[i].second << endl;
  }
}

int main(void)
{
  int m;
  cin >> N >> m;
  memset(visited,false,sizeof(visited));
  for(int i=0;i<m;i++)
  {
    int s,t;
    cin >> s>> t;
    G[s].push_back(t);
    G[t].push_back(s);
  }
  bridge_points();
}