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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using P = pair<int,int>;

struct Lowlink {
  const vector<vector<int>> &g;
  vector<int> ord, low;
  vector<int> articulation;
  vector<P> bridge;
  int k;

  Lowlink(const vector<vector<int>>& g) : g(g) {}

  virtual void dfs(int idx, int par) {
    ord[idx] = k++;
    low[idx] = ord[idx];
    bool is_articulation = false;
    int cnt = 0;
    for(auto to : g[idx]) {
      if(ord[to] < 0) {
        ++cnt;
        dfs(to,idx);
        low[idx] = min(low[idx],low[to]);
        if(par != -1 && low[to] >= ord[idx]) is_articulation = true;
        if(ord[idx] < low[to]) bridge.push_back(P(min(idx,to),max(idx,to)));
      } else if(to != par) {
        low[idx] = min(low[idx],ord[to]);
      }
    }
    if(par == -1 && cnt > 1) is_articulation = true;
    if(is_articulation) articulation.push_back(idx);
  }

  virtual void build() {
    ord.assign(g.size(),-1);
    low.assign(g.size(),0);
    k = 0;
    rep(i,g.size()) {
      if(ord[i] < 0) dfs(i,-1);
    }
  }
};

int main() {
  int n, m, u, v;
  cin >> n >> m;
  vector<vector<int>> g(n);
  rep(i,m) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  Lowlink llink(g);
  llink.build();
  sort(llink.bridge.begin(),llink.bridge.end());
  for(auto x : llink.bridge) cout << x.first << " " << x.second << endl;
}