Problem: 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>

Solution (in PYTHON3):

Solution:


# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def ri():
	return int(f.readline().strip())
def ria():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def rs():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]





from collections import deque
import math
# NUMBER = 10**9 + 7
NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
def choose(n,k):
	if n < k:
		assert false
	return  mult(factorial(n), modinv(mult(factorial(k),factorial(n-k))))
from collections import deque, defaultdict 
import heapq


def bfs(V, E, k):
	s = set()
	evens = set()
	odds = set()
	parents = defaultdict(int)
	depths = defaultdict(int)
	# print('\n'.join(map(str,E.items())))
	v = 1
	q = deque([])
	s.add(v)
	q.append((v,None,0))
	parents[1] = 0
	inds = (k+1)//2
	while len(q) > 0:
		x,parent,depth = q.popleft()
		if depth % 2 == 0:
			evens.add(x)
		else:
			odds.add(x)
		for u in E[x]:
			if u in s and u != parent:
				if depth + 1 + depths[u] <= k:
					return depths,parents,u,x
			if u not in s:
				s.add(u)
				q.append((u,x,depth+1))
				parents[u] = x
				depths[u] = depth+1
		if len(evens) >= inds:
			return (evens,)
		if len(odds) >= inds:
			return (odds,)
		# print(q)


def solution(V,E,n,m,k):
	res = bfs(V,E,k)
	if len(res) > 1:
		depths, parents, v,u = res
		u_path = []
		s = set()
		while u != 0:
			s.add(u)
			u_path.append(u)
			u = parents[u]
		# print(u_path)
		v_path = []
		while v != 0 and v not in s:
			# print(v)
			v_path.append(v)
			v = parents[v]
		u_path = list(reversed(u_path))
		for i in range(len(u_path)):
			if u_path[i] == v:
				break
		u_path = u_path[i:]
		sol = u_path + v_path
		return'2\n' + str(len(sol)) + '\n' +' '.join(map(str,sol))
	else:
		s = list(res[0])[:(k+1)//2]
		return '1\n'+' '.join(map(str,s))







	
def main():
	T = 1
	for i in range(T):
		# n = ri()
		n,m,k = ria()
		E = defaultdict(list)
		V = set()
		for _ in range(m):
			u,v = ria()
			E[u].append(v)
			E[v].append(u)
			V.add(u)
			V.add(v)
		# b = ria()

		x = solution(V,E,n,m,k)
		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()