#include "bits/stdc++.h"
using namespace std;

class Bridge {
public:
	vector<vector<int>> G;
	vector<int> ord, low;
	vector<pair<int, int>> bridges;
	int k;

	Bridge(int N):k(0) {
		G.resize(N); ord.resize(N, -1); low.resize(N, -1);
	}
	void clear() { G.clear();ord.clear();low.clear(); }
	void add_edge(int s, int t) {
		G[s].emplace_back(t); 
		G[t].emplace_back(s);
	}

	bool is_bridge(int u, int v)const {
		if (ord[u] > ord[v])swap(u, v);
		return ord[u] < low[v];
	}

	/* Lowlink */
	void dfs(int u,int rev = -1) {
		ord[u] = low[u] = k;
		k++;
		for (auto v : G[u]) {
			if (v == rev)continue;
			if (ord[v] < 0) {
				dfs(v, u);
				low[u] = min(low[u], low[v]);
			}
			else {
				low[u] = min(low[u], ord[v]);
			}
			if (is_bridge(u, v)) {
				int s = u, t = v;
				if (s > t) swap(s, t);
				bridges.push_back({ s,t });
			}
		}
	}

	void solve() {
		dfs(0);
		sort(bridges.begin(), bridges.end());
	}
};

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	int V, E; cin >> V >> E;
	Bridge b(V);
	for (int i = 0; i < E;i++) {
		int s, t; cin >> s >> t;
		b.add_edge(s, t);
	}
	b.solve();
	for (auto bridge : b.bridges) {
		cout << bridge.first << " " << bridge.second << endl;
	}
	return 0;
}
### Prompt

Please provide a CPP coded solution 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;

class Bridge {
public:
	vector<vector<int>> G;
	vector<int> ord, low;
	vector<pair<int, int>> bridges;
	int k;

	Bridge(int N):k(0) {
		G.resize(N); ord.resize(N, -1); low.resize(N, -1);
	}
	void clear() { G.clear();ord.clear();low.clear(); }
	void add_edge(int s, int t) {
		G[s].emplace_back(t); 
		G[t].emplace_back(s);
	}

	bool is_bridge(int u, int v)const {
		if (ord[u] > ord[v])swap(u, v);
		return ord[u] < low[v];
	}

	/* Lowlink */
	void dfs(int u,int rev = -1) {
		ord[u] = low[u] = k;
		k++;
		for (auto v : G[u]) {
			if (v == rev)continue;
			if (ord[v] < 0) {
				dfs(v, u);
				low[u] = min(low[u], low[v]);
			}
			else {
				low[u] = min(low[u], ord[v]);
			}
			if (is_bridge(u, v)) {
				int s = u, t = v;
				if (s > t) swap(s, t);
				bridges.push_back({ s,t });
			}
		}
	}

	void solve() {
		dfs(0);
		sort(bridges.begin(), bridges.end());
	}
};

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	int V, E; cin >> V >> E;
	Bridge b(V);
	for (int i = 0; i < E;i++) {
		int s, t; cin >> s >> t;
		b.add_edge(s, t);
	}
	b.solve();
	for (auto bridge : b.bridges) {
		cout << bridge.first << " " << bridge.second << endl;
	}
	return 0;
}
```