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<algorithm>
#include<vector>
#include<map>
#include<set>
#include<utility>

using namespace std;

using P = pair<int, int>;
int ord[100000];
int low[100000];
int v, e;

set<P> M;
vector<int> Edge[100000];
vector<bool> isNew(100000, true);
set<P> U;

bool isEdge(int s, int t) {
	return M.end()!=M.find(P(s, t));
}
bool isUsed(int s, int t) {
	return U.end()!=U.find(P(s, t));
}
void dfs(int u) {
	isNew[u] = false;
	static int k ;
	ord[u] = k; k++;
	//cout << "u=" << u << " k=" << k << endl;
	low[u] = ord[u];
	for (int v : Edge[u]) {
		if (isNew[v]) {
			U.emplace(u, v); U.emplace(v, u);
			dfs(v);
			low[u] = min(low[u], low[v]);
		}
		else if (!isUsed(u, v)) {
			low[u] = min(low[u], ord[v]);
		}
	}
}
int main() {
	cin >> v >> e;
	for (int i = 0; i < e; i++) {
		int s, t;
		cin >> s >> t;
		M.emplace(s, t);
		M.emplace(t, s);
		Edge[s].push_back(t);
		Edge[t].push_back(s);
	}
	dfs(0);
	//for (P p : M) {
	//	cout << p.first << " " << p.second << endl;
	//}
	//for (int i = 0; i < v; i++) {
	//	cout << "low=" << low[v] << " ord=" << ord[v] << endl;
	//}
	//cout << endl;
	for (P p : M) {
		int f = p.first;
		int s = p.second;
		if (f < s) {

		if (ord[f] < low[s] || ord[s] < low[f]) {
			cout << f << " " << p.second << endl;
		}
		}
	}
}