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;
using i64 = long long;
#define rep(i,s,e) for(int (i) = (s);(i) <= (e);(i)++)
#define all(x) x.begin(),x.end()

#include <vector>
#include <set>
#include <functional>
using namespace std;

vector<pair<int,int>> lowlink(const vector<vector<int>>& G){
  vector<pair<int,int>> bridge;
  //vector<int> articulation;
  int n = G.size();
  vector<int> used(n,false),ord(n,0),low(n,0);
  function<int(int,int,int)> dfs = [&](int v,int k,int f){
    used[v] = true;
    ord[v] = k++;
    low[v] = ord[v];
    int cnt = 0;
    bool is_articulation = false;
    for(auto & to : G[v]){
      if(!used[to]){
        cnt++;
        k = dfs(to,k,v);
        low[v] = min(low[v] , low[to]);
        is_articulation |= (f != -1) && low[to] >= ord[v]; 
        if(ord[v] < low[to]) bridge.push_back({min(v,to),max(v,to)});
      }
      else if(to != f){
        low[v] = min(low[v] , ord[to]);
      }
    }
    is_articulation |= f == -1 && cnt > 1;
    //if(is_articulation) articulation.push_back(v);
    return k;
  };
  dfs(0,0,-1);
  return bridge;
}

int main(){
  int N,M;
  cin >> N >> M;
  vector<vector<int>> G(N);

  rep(i,0,M - 1){
    int a,b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }

  auto ans = lowlink(G);

  sort(all(ans));
  for(auto p : ans){
    cout << p.first << " " << p.second << endl;
  }
}