#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
vector<vector<int>> G(100000);
vector<bool> visit(100000, false);
vector<int> prenum(100000), parent(100000, -1), lowest(100000);
int timer = 1;

void dfs(int u, int prev) {
	prenum[u] = lowest[u] = timer++;
	visit[u] = true;
	for (int v : G[u]) {
		if (!visit[v]) {
			parent[v] = u;
			dfs(v, u);
			lowest[u] = min(lowest[u], lowest[v]);
		}
		else if (v != prev) {
			lowest[u] = min(lowest[u], prenum[v]);
		}
	}
}

int main() {
	cin.tie(0);ios::sync_with_stdio(false);
	int V, E; cin >> V >> E;
	set<pair<int, int>> Ans;
	for (int i = 0; i < E;i++) {
		int s, t; cin >> s >> t;
		G[s].emplace_back(t);
		G[t].emplace_back(s);
	}
	dfs(0, -1);

	//u:??? v:???
	//prenum[u] < lowest[v]

	for (int i = 0; i < V;i++) {
		for (int v : G[i]) {
			if (prenum[i] < lowest[v]) {
				if (i <= v) { Ans.insert(make_pair(i, v)); }
				else { Ans.insert(make_pair(v, i)); }
			}
		}
	}

	for (auto p : Ans) {
		cout << p.first << " " << p.second << endl;
	}
}
### Prompt

Please create a solution in Cpp to the following 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

### Response

```cpp
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
vector<vector<int>> G(100000);
vector<bool> visit(100000, false);
vector<int> prenum(100000), parent(100000, -1), lowest(100000);
int timer = 1;

void dfs(int u, int prev) {
	prenum[u] = lowest[u] = timer++;
	visit[u] = true;
	for (int v : G[u]) {
		if (!visit[v]) {
			parent[v] = u;
			dfs(v, u);
			lowest[u] = min(lowest[u], lowest[v]);
		}
		else if (v != prev) {
			lowest[u] = min(lowest[u], prenum[v]);
		}
	}
}

int main() {
	cin.tie(0);ios::sync_with_stdio(false);
	int V, E; cin >> V >> E;
	set<pair<int, int>> Ans;
	for (int i = 0; i < E;i++) {
		int s, t; cin >> s >> t;
		G[s].emplace_back(t);
		G[t].emplace_back(s);
	}
	dfs(0, -1);

	//u:??? v:???
	//prenum[u] < lowest[v]

	for (int i = 0; i < V;i++) {
		for (int v : G[i]) {
			if (prenum[i] < lowest[v]) {
				if (i <= v) { Ans.insert(make_pair(i, v)); }
				else { Ans.insert(make_pair(v, i)); }
			}
		}
	}

	for (auto p : Ans) {
		cout << p.first << " " << p.second << endl;
	}
}
```