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;
using graph = vector<vector<int>>;

vector<int> biconnected_component(const graph& G) {
	int n = G.size(), idx = 0, t = 0, k = 0;
	vector<int> ord(n, -1), stk, rts(n), cmp(n);
	vector<bool> onS(n);
	function<void(int, int)> dfs = [&](int v, int u) {
		ord[v] = idx++;
		stk.push_back(v);
		onS[v] = true;
		rts[t++] = v;
		for (int w : G[v]) {
			if (ord[w] == -1) dfs(w, v);
			else if (u != w && onS[w])
				while (ord[rts[t - 1]] > ord[w]) --t;
		}
		if (v == rts[t - 1]) {
			while (true) {
				int w = stk.back(); stk.pop_back();
				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, -1);
	}
	return cmp;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0);
	int n, m;
	cin >> n >> m;
	graph G(n);
	vector<pair<int, int>> es;
	for (int i = 0; i < m; i++) {
		int s, t;
		cin >> s >> t;
		if (s > t) swap(s, t);
		G[s].push_back(t);
		G[t].push_back(s);
		es.emplace_back(s, t);
	}
	auto bc = biconnected_component(G);
	sort(es.begin(), es.end());
	for (auto& p : es) if (bc[p.first] != bc[p.second]) {
		printf("%d %d\n", p.first, p.second);
	}
	return 0;
}