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 <bits/stdc++.h>
using namespace std;

struct Edge {
	int from, to;
	Edge(int f, int t) : from(f), to(t) {}
};

using Edges = vector<Edge>;
using Graph = vector<Edges>;
pair<vector<int>, Edges> bridge(const Graph& g) {
	const int n = g.size();
	int idx = 0, s = 0, t = 0, k = 0;
	vector<int>ord(n, -1), onS(n), stk(n), roots(n), cmp(n);
	Edges brdg;
	function<void(int, int)> dfs = [&](int v, int u) {
		ord[v] = idx++;
		stk[s++] = v;
		onS[v] = true;
		roots[t++] = v;
		for (auto& e : g[v]) {
			int w = e.to;
			if (ord[w] == -1)dfs(w, v);
			else if (u != w && onS[w])
				while (ord[roots[t - 1]]>ord[w])--t;
		}
		if (v == roots[t - 1]) {
			brdg.emplace_back(u, v);
			while (true) {
				int w = stk[--s];
				onS[w] = false;
				cmp[w] = k;
				if (v == w)break;
			}
			--t;
			++k;
		}
	};
	for (int u = 0; u<n; ++u) {
		if (ord[u] == -1) {
			dfs(u, n);
			brdg.pop_back();
		}
	}
	return make_pair(cmp, brdg);
}

int main()
{
	int V, E;
	cin >> V >> E;
	Graph G(V);
	for (int i = 0, s, t; i < E; i++) {
		cin >> s >> t;
		G[s].emplace_back(s, t);
		G[t].emplace_back(t, s);
	}
	auto br = bridge(G);
	vector<pair<int, int>> res;
	for (auto& e : br.second) {
		res.emplace_back(min(e.from, e.to), max(e.from, e.to));
	}
	sort(res.begin(), res.end());
	for (auto p : res) {
		cout << p.first << ' ' << p.second << endl;
	}
	return 0;
}