Given a connected undirected graph with n vertices and an integer k, you have to either:

  * either find an independent set that has exactly ⌈k/2⌉ 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 ≤ k ≤ n ≤ 10^5, n-1 ≤ m ≤ 2 ⋅ 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 ≤ u,v ≤ 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 ⌈k/2⌉ 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:

<image>

Notice that printing the independent set \{1,3\} or printing the cycle 2-1-4 is also OK.

In the third sample:

<image>

In the fourth sample:

<image>
import sys
input = sys.stdin.buffer.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 vis[i]==0:
					queue.append(i)
					self.parent[i] = element
					vis[i] = 1
		return -1

	def reroot(self, root, ans):
		stack = [root]
		vis = [0]*n
		while len(stack)!=0:
			e = stack[-1]
			if vis[e]:
				stack.pop()
				# Reverse_The_Change()
				continue
			vis[e] = 1
			for i in graph[e]:
				if not vis[e]:
					stack.append(i)
			if self.parent[e]==-1:
				continue
			# Change_The_Answers()

	def dfss(self, root):

		vis=[0]*self.n
		color=[-1]*self.n
		color[root]=0
		# self.pdfs(root,vis,color)
		# return
		stack=[root]
		while len(stack)!=0:
			e=stack.pop()
			if vis[e]:
				continue

			if color[e]==-1:
				if color[self.parent[e]]:
					color[e]=0
					g1.append(e+1)
				else:
					color[e]=1
					g2.append(e+1)
			vis[e]=1

			for i in self.graph[e]:
				if not vis[i]:
					stack.append(i)
					self.parent[i]=e

		return

n,m,k = map(int,input().split())
g = Graph(n,False)
for i in range(m):
	a,b = map(int,input().split())
	g.addEdge(a-1,b-1)
l = g.ifcycle()
# print (l)
if l!=-1 and len(l)<=k:
	print (2)
	print (len(l))
	for i in range(len(l)):
		l[i] += 1
	print (*l)
	exit()
g1,g2 = [1],[]
g.dfss(0)
if len(g1)>=(k-1)//2+1:
	print (1)
	print (*g1[0:(k-1)//2+1])
else:
	print (1)
	print (*g2[0:(k-1)//2+1])
