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.