Underground Lab
Contest: Технокубок 2017 - Финал (только для онсайт-финалистов)
Difficulty rating: 2100
Tags: constructive algorithms, dfs and similar, graphs

The evil Bumbershoot corporation produces clones for gruesome experiments in a vast underground lab. On one occasion, the corp cloned a boy Andryusha who was smarter than his comrades. Immediately Andryusha understood that something fishy was going on there. He rallied fellow clones to go on a feud against the evil corp, and they set out to find an exit from the lab. The corp had to reduce to destroy the lab complex.

The lab can be pictured as a connected graph with n vertices and m edges. k clones of Andryusha start looking for an exit in some of the vertices. Each clone can traverse any edge once per second. Any number of clones are allowed to be at any vertex simultaneously. Each clone is allowed to stop looking at any time moment, but he must look at his starting vertex at least. The exit can be located at any vertex of the lab, hence each vertex must be visited by at least one clone.

Each clone can visit at most $$\left\lceil \frac{2n}{k} \right\rceil$$ vertices before the lab explodes.

Your task is to choose starting vertices and searching routes for the clones. Each route can have at most $$\left\lceil \frac{2n}{k} \right\rceil$$ vertices.

Input format:
The first line contains three integers n, m, and k (1 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105, 1 ≤ k ≤ n) — the number of vertices and edges in the lab, and the number of clones.

Each of the next m lines contains two integers xi and yi (1 ≤ xi, yi ≤ n) — indices of vertices connected by the respective edge. The graph is allowed to have self-loops and multiple edges.

The graph is guaranteed to be connected.

Output format:
You should print k lines. i-th of these lines must start with an integer ci ($$1 \leq c_i \leq \left\lceil \frac{2n}{k} \right\rceil$$) — the number of vertices visited by i-th clone, followed by ci integers — indices of vertices visited by this clone in the order of visiting. You have to print each vertex every time it is visited, regardless if it was visited earlier or not.

It is guaranteed that a valid answer exists.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first sample case there is only one clone who may visit vertices in order (2, 1, 3), which fits the constraint of 6 vertices per clone.

In the second sample case the two clones can visited vertices in order (2, 1, 3) and (4, 1, 5), which fits the constraint of 5 vertices per clone.

Example input:
3 2 1
2 1
3 1

Example output:
3 2 1 3

Example input:
5 4 2
1 2
1 3
1 4
1 5

Example output:
3 2 1 3
3 4 1 5

Editorial:
Let's start a DFS at any vertex of the graph, and produce an Euler tour — the order of vertices visited by a DFS, where each vertex v is written down every time DFS visits it (in particular, when a recursive call made from v terminates). Note that the Euler tour has exactly 2n - 1 entries in it, hence it would be a correct answer for k = 1. For a general k, cut the Euler tour into k consecutive pieces of size at most ⌈ 2n / k⌉, and yield it as an answer. Note that each path of the answer has to contain at least one vertex.
Complexity: O(n + m) time and memory.