#include <bits/stdc++.h>
#define fi first
#define se second
using namespace std;
typedef pair <int, int> pii;

const int INF = 0x3f3f3f3f;
int n, e, t1, t2, timer, st[100005], nd[100005], low[100005];
vector <int> v[100005];
vector <pii> b;

void dfs(int node, int p) {
	st[node] = timer++;
	low[node] = st[node];
	for (int i : v[node]) {
		if (i == p) continue;
		if (st[i] == INF) {
			dfs(i, node);
			low[node] = min(low[node], low[i]);
			if (low[i] > st[node]) {
				b.push_back(make_pair(i, node));
			}
		}
		else if (nd[i] == INF) {
			low[node] = min(low[node], low[i]);
		}
	}
	nd[node] = timer;
}

int main() {
	cin >> n >> e;
	for (int i = 0; i < n; i++) st[i] = INF, nd[i] = INF;
	for (int i = 0; i < e; i++) {
		cin >> t1 >> t2;
		v[t1].push_back(t2);
		v[t2].push_back(t1);
	}
	dfs(0, -1);
//	for (int i = 0; i < n; i++) {
//		cout << st[i] << ' ' << nd[i] << ' ' << low[i] << endl;
//	}
	for (int i = 0; i < b.size(); i++) if (b[i].fi > b[i].se) swap(b[i].fi, b[i].se);
	sort(b.begin(), b.end());
	for (int i = 0; i < b.size(); i++) cout << b[i].fi << ' ' << b[i].se << endl;
	return 0;
}



### Prompt

Please formulate a CPP solution 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 <bits/stdc++.h>
#define fi first
#define se second
using namespace std;
typedef pair <int, int> pii;

const int INF = 0x3f3f3f3f;
int n, e, t1, t2, timer, st[100005], nd[100005], low[100005];
vector <int> v[100005];
vector <pii> b;

void dfs(int node, int p) {
	st[node] = timer++;
	low[node] = st[node];
	for (int i : v[node]) {
		if (i == p) continue;
		if (st[i] == INF) {
			dfs(i, node);
			low[node] = min(low[node], low[i]);
			if (low[i] > st[node]) {
				b.push_back(make_pair(i, node));
			}
		}
		else if (nd[i] == INF) {
			low[node] = min(low[node], low[i]);
		}
	}
	nd[node] = timer;
}

int main() {
	cin >> n >> e;
	for (int i = 0; i < n; i++) st[i] = INF, nd[i] = INF;
	for (int i = 0; i < e; i++) {
		cin >> t1 >> t2;
		v[t1].push_back(t2);
		v[t2].push_back(t1);
	}
	dfs(0, -1);
//	for (int i = 0; i < n; i++) {
//		cout << st[i] << ' ' << nd[i] << ' ' << low[i] << endl;
//	}
	for (int i = 0; i < b.size(); i++) if (b[i].fi > b[i].se) swap(b[i].fi, b[i].se);
	sort(b.begin(), b.end());
	for (int i = 0; i < b.size(); i++) cout << b[i].fi << ' ' << b[i].se << endl;
	return 0;
}



```