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;
vector<vector<int>> grh;
vector<int> ord, low;
vector<pair<int, int>> brg;
int cnt;
void dfs (int crr, int prt) {
  low[crr] = ord[crr] = cnt++;
  for (int nxt : grh[crr]) {
    if (ord[nxt] == -1) {
      dfs(nxt, crr);
      if (ord[nxt] == low[nxt]) brg.push_back({crr, nxt});
      low[crr] = min(low[crr], low[nxt]);
    } else if (nxt != prt) {
      low[crr] = min(low[crr], ord[nxt]);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  grh.resize(n);
  ord.resize(n, -1);
  low.resize(n);
  for (int i = 0; i < m; i++) {
    int s, t;
    cin >> s >> t;
    grh[s].push_back(t);
    grh[t].push_back(s);
  }
  cnt = 0;
  dfs(0, 0);
  for (auto& p : brg) {
    int& crr = p.first;
    int& nxt = p.second;
    if (crr > nxt) swap(crr, nxt);
  }
  sort(brg.begin(), brg.end());
  for (auto p : brg) {
    int crr = p.first;
    int nxt = p.second;
    cout << crr << ' ' << nxt << '\n';
  }
  return 0;
}