// 参考:http://kagamiz.hatenablog.com/entry/2013/10/05/005213

#include <stdio.h>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> G[1000000];
vector<pair<int, int> > bridge;
int ord[1000000], low[1000000];
bool vis[1000000];

void dfs(int v, int p, int &k){
	vis[v] = true;
	ord[v] = k++;
	low[v] = ord[v];
	int ct = 0;
	
	for (int i = 0; i < G[v].size(); i++){
		if (!vis[G[v][i]]){
			ct++;
			dfs(G[v][i], v, k);
			low[v] = min(low[v], low[G[v][i]]);
			if (ord[v] < low[G[v][i]]) bridge.push_back(make_pair(min(v, G[v][i]), max(v, G[v][i])));
		}
		else if (G[v][i] != p){
			low[v] = min(low[v], ord[G[v][i]]);
		}
	}
}

int main()
{
	int n,m;
	scanf("%d %d", &n, &m);
	
	for (int i = 0;i < m;++i){
		int v, e;
		scanf("%d %d",&v,&e);
		G[v].push_back(e);
		G[e].push_back(v);
	}
	int k = 0;
	for (int i = 0; i < n; i++){
		if (!vis[i]) dfs(i, -1, k);
	}
	
	sort(bridge.begin(), bridge.end());
	for (int i = 0; i < bridge.size(); i++)
		printf("%d %d\n", bridge[i].first, bridge[i].second);
}

### 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
// 参考:http://kagamiz.hatenablog.com/entry/2013/10/05/005213

#include <stdio.h>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> G[1000000];
vector<pair<int, int> > bridge;
int ord[1000000], low[1000000];
bool vis[1000000];

void dfs(int v, int p, int &k){
	vis[v] = true;
	ord[v] = k++;
	low[v] = ord[v];
	int ct = 0;
	
	for (int i = 0; i < G[v].size(); i++){
		if (!vis[G[v][i]]){
			ct++;
			dfs(G[v][i], v, k);
			low[v] = min(low[v], low[G[v][i]]);
			if (ord[v] < low[G[v][i]]) bridge.push_back(make_pair(min(v, G[v][i]), max(v, G[v][i])));
		}
		else if (G[v][i] != p){
			low[v] = min(low[v], ord[G[v][i]]);
		}
	}
}

int main()
{
	int n,m;
	scanf("%d %d", &n, &m);
	
	for (int i = 0;i < m;++i){
		int v, e;
		scanf("%d %d",&v,&e);
		G[v].push_back(e);
		G[e].push_back(v);
	}
	int k = 0;
	for (int i = 0; i < n; i++){
		if (!vis[i]) dfs(i, -1, k);
	}
	
	sort(bridge.begin(), bridge.end());
	for (int i = 0; i < bridge.size(); i++)
		printf("%d %d\n", bridge[i].first, bridge[i].second);
}

```