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;

const int MAX_V = 1e5+10;

vector<vector<int> > g(MAX_V);
vector<pair<int, int> > bridge;
int n;
bool visited[MAX_V];
int ord[MAX_V], low[MAX_V], timer;

void dfs(int v, int p){
	ord[v] = low[v] = timer++;
	visited[v] = true;
	for (int i = 0; i < g[v].size(); ++i)
	{
		int to = g[v][i];
		if (to == p) continue;
		if (visited[to]) low[v] = min(low[v], ord[to]);
		else {
			dfs(to, v);
			low[v] = min(low[v], low[to]);
			if (ord[v] < low[to]) bridge.push_back(make_pair(min(v,to),max(v, to)));
		}
	}
}

void find_bridge(){
	for (int i = 0; i < n; ++i) visited[i] = false;

	timer = 0;
    dfs(0,-1);
    sort(bridge.begin(), bridge.end());
	bridge.erase(unique(bridge.begin(), bridge.end()), bridge.end());
}

int main(int argc, char const *argv[])
{
	int m;
	cin >> n >> m;
	for (int i = 0; i < m; ++i)
	{
		int s, t;
		cin >> s >> t;
		g[s].push_back(t);
		g[t].push_back(s);
	}

	find_bridge();

	for (int i = 0; i < bridge.size(); ++i) cout << bridge[i].first << ' ' << bridge[i].second << endl;
	return 0;
}