#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<functional>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<numeric>
#include<limits>
#include<iterator>

#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define rep(i,n) for(int i=0; i<n; i++)
#define FOR(i,a,n) for(int i=a; i<n; i++)

using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ll, char> plc;
typedef vector<int> vi;
typedef vector<vi> vvi;

int dfs(vvi& g, vector<pi>& res, int v, int& count, int from, vi& low, vi& pre)
{
	pre[v] = count++;
	low[v] = pre[v];
	for (vi::iterator it = g[v].begin(); it != g[v].end(); it++) {
		int to = *it;
		if (pre[to] == -1) {
		
			low[v] = min(low[v], dfs(g, res, to, count, v, low, pre));
			if (low[to] == pre[to]) {
				res.push_back(make_pair(min(to,v), max(to,v)));
			}
		}
		else {
			if (from == to) {
				continue;
			}
			low[v] = min(low[v], low[to]);
		}
	}
	return low[v];
}

// g: adjacency list
// v:number of nodes

vector<pi> bridges(vvi& g, int v) {
	vector<pi> res;
	if (v > 0) {
		
		vi low(v, -1); 
		vi pre(v, -1); 
		int count = 0; 
		dfs(g, res, 0, count, -1, low, pre); 
	}
	return res;
}

vvi G;
vector<pi> ans;

int main()
{
	int v, e;
	cin >> v >> e;
	G.resize(v);

	rep(i, e) {
		int s, t;
		cin >> s >> t;
		G[s].push_back(t);
		G[t].push_back(s);
	}

	ans = bridges(G, v);
	sort(ans.begin(), ans.end());
	
	rep(i, ans.size())
		cout << ans[i].first << " " << ans[i].second << endl;
	return 0;
}
### 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<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<functional>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<numeric>
#include<limits>
#include<iterator>

#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define rep(i,n) for(int i=0; i<n; i++)
#define FOR(i,a,n) for(int i=a; i<n; i++)

using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ll, char> plc;
typedef vector<int> vi;
typedef vector<vi> vvi;

int dfs(vvi& g, vector<pi>& res, int v, int& count, int from, vi& low, vi& pre)
{
	pre[v] = count++;
	low[v] = pre[v];
	for (vi::iterator it = g[v].begin(); it != g[v].end(); it++) {
		int to = *it;
		if (pre[to] == -1) {
		
			low[v] = min(low[v], dfs(g, res, to, count, v, low, pre));
			if (low[to] == pre[to]) {
				res.push_back(make_pair(min(to,v), max(to,v)));
			}
		}
		else {
			if (from == to) {
				continue;
			}
			low[v] = min(low[v], low[to]);
		}
	}
	return low[v];
}

// g: adjacency list
// v:number of nodes

vector<pi> bridges(vvi& g, int v) {
	vector<pi> res;
	if (v > 0) {
		
		vi low(v, -1); 
		vi pre(v, -1); 
		int count = 0; 
		dfs(g, res, 0, count, -1, low, pre); 
	}
	return res;
}

vvi G;
vector<pi> ans;

int main()
{
	int v, e;
	cin >> v >> e;
	G.resize(v);

	rep(i, e) {
		int s, t;
		cin >> s >> t;
		G[s].push_back(t);
		G[t].push_back(s);
	}

	ans = bridges(G, v);
	sort(ans.begin(), ans.end());
	
	rep(i, ans.size())
		cout << ans[i].first << " " << ans[i].second << endl;
	return 0;
}
```