#include <bits/stdc++.h>
#define pb push_back
using namespace std;

const int mx = 100005;

int dtime = 0;
int n, m, par[mx], low[mx], start[mx];
vector<int> adj[mx];
bool vis[mx];
vector<pair<int, int>> bdg;

void dfs(int n, int p = -1) {
  vis[n] = true;
  par[n] = p;
  start[n] = dtime++;
  low[n] = start[n];
  
  for (int h : adj[n]) {
    if (!vis[h]) {
      dfs(h, n);
      // printf("Low[%d] = min(%d, %d) = %d\n", n, low[n], low[h], min(low[n], low[h]));
      low[n] = min(low[h], low[n]);
      if (low[h] > start[n]) {
	bdg.push_back({min(n, h), max(n, h)});
      }
    } else if (h != p) {
      low[n] = min(low[n], start[h]);
    }
  }
}

int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v; cin >> u >> v;
    adj[u].pb(v); adj[v].pb(u);
  }
  dfs(0);
  sort(bdg.begin(), bdg.end());
  for (auto p : bdg) cout << p.first << " " << p.second << endl;
}


### Prompt

Construct a CPP code solution to the problem outlined:

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>
#define pb push_back
using namespace std;

const int mx = 100005;

int dtime = 0;
int n, m, par[mx], low[mx], start[mx];
vector<int> adj[mx];
bool vis[mx];
vector<pair<int, int>> bdg;

void dfs(int n, int p = -1) {
  vis[n] = true;
  par[n] = p;
  start[n] = dtime++;
  low[n] = start[n];
  
  for (int h : adj[n]) {
    if (!vis[h]) {
      dfs(h, n);
      // printf("Low[%d] = min(%d, %d) = %d\n", n, low[n], low[h], min(low[n], low[h]));
      low[n] = min(low[h], low[n]);
      if (low[h] > start[n]) {
	bdg.push_back({min(n, h), max(n, h)});
      }
    } else if (h != p) {
      low[n] = min(low[n], start[h]);
    }
  }
}

int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v; cin >> u >> v;
    adj[u].pb(v); adj[v].pb(u);
  }
  dfs(0);
  sort(bdg.begin(), bdg.end());
  for (auto p : bdg) cout << p.first << " " << p.second << endl;
}


```