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;

struct Graph {
  size_t n;
  int cnt;
  vector<vector<int>> g;
  vector<bool> isArt;
  vector<int> ord, low;
  vector<pair<int, int>> bridges;

  Graph(size_t n) : n(n), cnt(0), g(n), ord(n, -1), low(n, -1), isArt(n, false) {}

  void addEdge(int u, int v) {
    g[u].push_back(v);
    g[v].push_back(u);
  }
  void dfs(int u, int p) {
    ord[u] = low[u] = cnt++;
    for (auto&& v : g[u]) {
      if (ord[v] == -1) {
        dfs(v, u);
        low[u] = min(low[u], low[v]);
        if (ord[u] <= low[v]) isArt[u] = (ord[u] > 0) || (ord[v] > 1);
        if (ord[u] < low[v]) bridges.push_back({min(u, v), max(u, v)});
      } else if (v != p) {
        low[u] = min(low[u], ord[v]);
      }
    }
  }
  void biconnectedComponents() {
    for (int i = 0; i < static_cast<int>(n); ++i) {
      if (ord[i] == -1) dfs(i, -1);
    }
  }
};

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, m; cin >> n >> m;
  Graph g(n);
  while (m--) {
    int s, t; cin >> s >> t;
    g.addEdge(s, t);
  }
  g.biconnectedComponents();

  sort(begin(g.bridges), end(g.bridges));
  for (auto&& p : g.bridges) {
    cout << p.first << ' ' << p.second << endl;
  }
  return 0;
}