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 V_MAX 100000
using namespace std;
struct edgeB{int u,v;}; //u,v???????????????
int V;                  //???????????°
vector <int> G[V_MAX];  //??°???????????£??\??????????????¨???
vector <edgeB> bridges; // ???
int ord[V_MAX];         //?????????????????????????????????
int low[V_MAX];         //
 
void dfs(int u, int p, int &c){
  ord[u] = low[u] = c++;
  for(int i=0;i<G[u].size();i++){
    int v = G[u][i];
    if(v==p)continue;
    
    if(ord[v] == -1){
      dfs(v,u,c);
      low[u] = min(low[u],low[v]);
    }
    else low[u] = min(low[u], ord[v]);
    if(ord[u] < low[v]) bridges.push_back((edgeB){u,v});
  }
}

void bridge(){
  bridges.clear();
  int c = 0;
  memset(ord,-1,sizeof(ord));
  for(int i=0;i<V;i++) if(ord[i]==-1) dfs(i,-1,c);
}

bool compare(edgeB a, edgeB b){
  return a.u==b.u? a.v<b.v:a.u<b.u;
}

int main(){
  int E;
  cin>>V>>E;
  for(int i=0;i<E;i++){
    int s,t;
    cin>>s>>t;
    G[s].push_back(t);
    G[t].push_back(s);
  }
  
  bridge();
  for(int i=0;i<bridges.size();i++)
    if(bridges[i].u > bridges[i].v)swap(bridges[i].u, bridges[i].v);
  sort(bridges.begin(),bridges.end(),compare);
  for(int i=0;i<bridges.size();i++) cout << bridges[i].u<<" "<<bridges[i].v<<endl;
  return 0;
}