A13300 | Ehab's Last Theorem

It's the year 5555. You have a graph, and you want to find a long cycle and a huge independent set, just because you can. But for now, let's just stick with finding either.

Given a connected graph with $n$ vertices, you can choose to either:

- find an independent set that has exactly $\lceil\sqrt{n}\rceil$ vertices.
- find a simple cycle of length at least $\lceil\sqrt{n}\rceil$ .

An independent set is a set of vertices such that no two of them are connected by an edge. A simple cycle is a cycle that doesn't contain any vertex twice. I have a proof you can always solve one of these problems, but it's too long to fit this margin.

The first line contains two integers $n$ and $m$ ( $5 \le n \le 10^5$ , $n-1 \le m \le 2 \cdot 10^5$ ) — the number of vertices and edges in the graph.

Each of the next $m$ lines contains two space-separated integers $u$ and $v$ ( $1 \le u,v \le n$ ) that mean there's an edge between vertices $u$ and $v$ . It's guaranteed that the graph is connected and doesn't contain any self-loops or multiple edges.

If you choose to solve the first problem, then on the first line print "1", followed by a line containing $\lceil\sqrt{n}\rceil$ distinct integers not exceeding $n$ , the vertices in the desired independent set.

If you, however, choose to solve the second problem, then on the first line print "2", followed by a line containing one integer, $c$ , representing the length of the found cycle, followed by a line containing $c$ distinct integers integers not exceeding $n$ , the vertices in the desired cycle, in the order they appear in the cycle.