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 LowLink {
  using A = vector<int>;
  using B = vector<pair<int, int>>;
  int n;
  vector<vector<int>> es;
  vector<bool> vis;
  vector<int> ord, low;
  A articulation;
  B bridge;
  LowLink(int n) : n(n), es(n) {}
  void add_edge(int v, int u) {
    es[v].push_back(u);
    es[u].push_back(v);
  }
  void dfs(int v, int p, int &k) {
    vis[v] = true;
    ord[v] = k++;
    low[v] = ord[v];
    bool is = false;
    int cnt = 0;
    for (auto u : es[v]) {
      if (vis[u])  {
        if (u != p) low[v] = min(low[v], ord[u]);
        continue;
      }
      cnt++;
      dfs(u, v, k);
      low[v] = min(low[v], low[u]);
      if (p != -1 && ord[v] <= low[u]) is = true;
      if (ord[v] < low[u]) bridge.emplace_back(min(v, u), max(v, u));
    }
    if (p == -1 && cnt > 1) is = true;
    if (is) articulation.push_back(v);
  }
  pair<A, B> build() {
    int k = 0;
    vis.assign(n, false);
    ord.assign(n, 0);
    low.assign(n, 0);
    articulation.clear();
    bridge.clear();
    dfs(0, -1, k);
    return pair<A, B>(articulation, bridge);
  }
};

int main() {
  int N, M; cin >> N >> M;
  auto low = LowLink(N);
  for (int i = 0; i < M; i++) {
    int v, u; cin >> v >> u;
    low.add_edge(v, u);
  }
  auto ans = low.build().second;
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << '\n';
  }
  return 0;
}