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 <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	int V, E;
	cin >> V >> E;
	vector<vector<int> > G(V);
	for(int i = 0; i < E; ++i) {
		int a, b;
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	vector<int> dp(V, 1 << 30), par(V), depth(V);
	vector<bool> vis(V, false);
	function<void(int, int)> dfs = [&](int pos, int pre) {
		par[pos] = pre;
		vis[pos] = true;
		dp[pos] = depth[pos];
		for(int i : G[pos]) {
			if(i == pre) continue;
			if(!vis[i]) {
				depth[i] = depth[pos] + 1;
				dfs(i, pos);
				dp[pos] = min(dp[pos], dp[i]);
			}
			else {
				dp[pos] = min(dp[pos], depth[i]);
			}
		}
	};
	for(int i = 0; i < V; ++i) {
		if(!vis[i]) {
			dfs(i, -1);
		}
	}
	vector<pair<int, int> > bridges;
	for(int i = 0; i < V; ++i) {
		if(par[i] != -1) {
			if(dp[i] == depth[i]) {
				int a = i, b = par[i];
				if(a > b) swap(a, b);
				bridges.push_back(make_pair(a, b));
			}
		}
	}
	sort(bridges.begin(), bridges.end());
	for(pair<int, int> i : bridges) {
		cout << i.first << ' ' << i.second << '\n';
	}
	return 0;
}