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<set>
#include<algorithm>

using namespace std;

struct Graph{
  vector< vector<int> > edge;
  vector<bool> visited;
  vector<int> lowest, parent, prenum;
  int timer;

  Graph(int n) : edge(n), visited(n, false), lowest(n), parent(n, -1), prenum(n){
    timer = 1;
  }

  void dfs(int now, int prev){
    lowest[now] = prenum[now] = timer;
    timer++;
    visited[now] = true;

    for(auto next:edge[now]){
      if(!visited[next]){ //next = 子供
        parent[next] = now;
        dfs(next, now);
        lowest[now] = min(lowest[now], lowest[next]);
      }
      else if(next != prev){ //BackEdge next = 親戚
        lowest[now] = min(lowest[now], prenum[next]);
      }
    }
  }

  void bridge(){
    dfs(0,1);

    set< pair<int, int> > bridge;
    for(int i = 0; i < prenum.size(); i++){
      if(i == 0) continue;
      if(prenum[parent[i]] < lowest[i]) bridge.insert({min(i, parent[i]), max(i, parent[i])});
    }
    for(auto b:bridge){
      cout << b.first << " " << b.second << endl;
    }
  }
};

int main(){
  int v, e;
  cin >> v >> e;
  Graph G = (v);
  for(int i = 0; i < e; i++){
    int s,t;
    cin >> s >> t;
    G.edge[s].push_back(t);
    G.edge[t].push_back(s);
  }
  G.bridge();
  return 0;
}