#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int> g[300005];
bitset<300005> vis;
int tin[300005], low[300005], timer;
set<pair<int, int>> wtf;
void dfs(int u, int p = -1) {
  vis[u] = 1;
  tin[u] = low[u] = timer++;
  for (int v : g[u]) {
    if (v == p)
      continue;
    if (vis[v]) {
      low[u] = min(low[u], tin[v]);
    } else {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] > tin[u]) {
        wtf.insert({min(u, v), max(u, v)});
      }
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++)
    if (!vis[i])
      dfs(i);
  for (pair<int, int> i : wtf) {
    cout << i.first << ' ' << i.second << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int> g[300005];
bitset<300005> vis;
int tin[300005], low[300005], timer;
set<pair<int, int>> wtf;
void dfs(int u, int p = -1) {
  vis[u] = 1;
  tin[u] = low[u] = timer++;
  for (int v : g[u]) {
    if (v == p)
      continue;
    if (vis[v]) {
      low[u] = min(low[u], tin[v]);
    } else {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] > tin[u]) {
        wtf.insert({min(u, v), max(u, v)});
      }
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < n; i++)
    if (!vis[i])
      dfs(i);
  for (pair<int, int> i : wtf) {
    cout << i.first << ' ' << i.second << endl;
  }
}

```