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 collections import defaultdict

# f = """5638 77083 90733 13461 55247 51690 2764 98093 9714 59034 40460 88111 26002 2307 68646 47771 35240 77405 50249 99661 78405 26655 37140 25922 83482 76578 13653 91798 67304 78492 40724 82850 18640 87242 29895 24159 96314 50129 2123 3752 42955 96564 33037 27637 30217 56344 74511 11229 51625 98787 17591 33395 74987 64540 54287 93454 55121 74344 54166 52447 14087 1 14087 52447 54166 74344 55121 73337 83927 29528 82868
# """
# print(len(f.strip().split()))
# print(len(set(f.strip().split())))
# ddd = Counter(f.strip().split())
# print(ddd.most_common(5))

# cycle =  list(range(123))
# print(*cycle[:2 * desreslen:2])
# if len(cycle) <= k:
#     print(2)
#     print(len(cycle))
#     print(*cycle)
# else:
#     print(1)
#     print(*cycle[:2 * desreslen:2])
n, m, k = list(map(int, input().split()))
desreslen = (k // 2) + (k % 2)


links = defaultdict(set)
for v in range(m):
    a, b = list(map(int, input().split()))
    links[a].add(b)
    links[b].add(a)

label = defaultdict(int)
prev = defaultdict(int)

start = 0
stk = [1]
label[stk[start]] = 1

cycled = False

while start < len(stk):
    cur = stk[start]
    for v in links[cur]:
        if prev[cur] == v:
            continue
        if label[v] == 0:
            stk.append(v)
            label[v] = label[cur] + 1
            prev[v] = cur
        else:
            cycled = (cur, v)
            break
    if cycled:
        break

    start += 1

if cycled:
    a,b = cycled
    acycle= [a]
    bcycle = [b]
    if label[acycle[-1]] > label[bcycle[-1]]:
        acycle.append(prev[acycle[-1]])
    elif label[acycle[-1]] < label[bcycle[-1]]:
        bcycle.append(prev[bcycle[-1]])

    while acycle[-1] != bcycle[-1]:
        acycle.append(prev[acycle[-1]])
        bcycle.append(prev[bcycle[-1]])

    cycle = acycle[:-1] + bcycle[::-1]
    # print('cycle found:', cycle, acycle,bcycle)

    if len(cycle) <= k:
        print(2)
        print(len(cycle))
        print(*cycle)
    else:
        print(1)
        print(*cycle[:2*desreslen:2])
else:
    evens = [v for v in label if label[v] % 2 ]
    oddss = [v for v in label if (label[v] % 2 == 0)]
    if len(oddss)>len(evens):
        res = oddss
    else:
        res = evens
    print(1)
    print(*(list(res)[:desreslen]))

"""
7 9 3
1 7
7 6
6 5
5 2
2 3
3 4
4 5
5 3
2 4

"""