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

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int ord[110000], low[110000], n, m, a, b, cnt;
vector<int>x[110000]; bool used[110000];
vector<pair<int, int>>bridge;
void dfs(int v, int k) {
	ord[v] = cnt; used[v] = true;
	low[v] = cnt; cnt++;
	for (int i = 0; i < x[v].size(); i++) {
		if (used[x[v][i]] == false) {
			dfs(x[v][i], v);
			low[v] = min(low[v], low[x[v][i]]);
			if (ord[v] < low[x[v][i]]) { bridge.push_back(make_pair(v, x[v][i])); }
		}
		else if (x[v][i] != k) {
			low[v] = min(low[v], ord[x[v][i]]);
		}
	}
}
int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> a >> b;
		x[a].push_back(b);
		x[b].push_back(a);
	}
	used[1] = true; dfs(1, -1);
	for (int i = 0; i < bridge.size(); i++) {
		if (bridge[i].first > bridge[i].second) { swap(bridge[i].first, bridge[i].second); }
	}
	sort(bridge.begin(), bridge.end());
	for (int i = 0; i < bridge.size(); i++) {
		cout << bridge[i].first << ' ' << bridge[i].second << endl;
	}
	return 0;
}