inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1364/D:
Given a connected undirected graph with $n$ vertices and an integer $k$, you have to either:  either find an independent set that has exactly $\lceil\frac{k}{2}\rceil$ vertices. or find a simple cycle of length at most $k$. 

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 that for any input you can always solve at least one of these problems, but it's left as an exercise for the reader.


-----Input-----

The first line contains three integers $n$, $m$, and $k$ ($3 \le k \le n \le 10^5$, $n-1 \le m \le 2 \cdot 10^5$) — the number of vertices and edges in the graph, and the parameter $k$ from the statement.

Each of the next $m$ lines contains two 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.


-----Output-----

If you choose to solve the first problem, then on the first line print $1$, followed by a line containing $\lceil\frac{k}{2}\rceil$ 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 not exceeding $n$, the vertices in the desired cycle, in the order they appear in the cycle.


-----Examples-----
Input
4 4 3
1 2
2 3
3 4
4 1

Output
1
1 3 
Input
4 5 3
1 2
2 3
3 4
4 1
2 4

Output
2
3
2 3 4 
Input
4 6 3
1 2
2 3
3 4
4 1
1 3
2 4

Output
2
3
1 2 3 
Input
5 4 5
1 2
1 3
2 4
2 5

Output
1
1 4 5 


-----Note-----

In the first sample:

[Image]

Notice that printing the independent set $\{2,4\}$ is also OK, but printing the cycle $1-2-3-4$ isn't, because its length must be at most $3$.

In the second sample:

$N$

Notice that printing the independent set $\{1,3\}$ or printing the cycle...

I tried it in Python, but could not do it. Can you solve it?
targets: import sys
input = sys.stdin.readline
from collections import deque

class Graph(object):
	"""docstring for Graph"""
	def __init__(self,n,d): # Number of nodes and d is True if directed
		self.n = n
		self.graph = [[] for i in range(n)]
		self.parent = [-1 for i in range(n)]
		self.directed = d
		
	def addEdge(self,x,y):
		self.graph[x].append(y)
		if not self.directed:
			self.graph[y].append(x)

	def bfs(self, root): # NORMAL BFS
		queue = [root]
		queue = deque(queue)
		vis = [0]*self.n
		while len(queue)!=0:
			element = queue.popleft()
			vis[element] = 1
			for i in self.graph[element]:
				if vis[i]==0:
					queue.append(i)
					self.parent[i] = element
					vis[i] = 1

	def dfs(self, root, ans): # Iterative DFS
		stack=[root]
		vis=[0]*self.n
		stack2=[]
		while len(stack)!=0: # INITIAL TRAVERSAL
			element = stack.pop()
			if vis[element]:
				continue
			vis[element] = 1
			stack2.append(element)
			for i in self.graph[element]:
				if vis[i]==0:
					self.parent[i] = element
					stack.append(i)

		while len(stack2)!=0: # BACKTRACING. Modify the loop according to the question
			element = stack2.pop()
			m = 0
			for i in self.graph[element]:
				if i!=self.parent[element]:
					m += ans[i]
			ans[element] = m
		return ans

	def shortestpath(self, source, dest): # Calculate Shortest Path between two nodes
		self.bfs(source)
		path = [dest]
		while self.parent[path[-1]]!=-1:
			path.append(parent[path[-1]])
		return path[::-1]

	def ifcycle(self):
		queue = [0]
		vis = [0]*n
		queue = deque(queue)
		while len(queue)!=0:
			element = queue.popleft()
			vis[element] = 1
			for i in self.graph[element]:
				if vis[i]==1 and i!=self.parent[element]:
					s = i
					e = element
					path1 = [s]
					path2 = [e]
					while self.parent[s]!=-1:
						s = self.parent[s]
						path1.append(s)
					while self.parent[e]!=-1:
						e = self.parent[e]
						path2.append(e)
					for i in range(-1,max(-len(path1),-len(path2))-1,-1):
						if path1[i]!=path2[i]:
							return path1[0:i+1]+path2[i+1::-1]
				if...