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
import math
from collections import defaultdict
from collections import deque

N, M, K = [int(_) for _ in sys.stdin.readline().split()]
edges = [set() for i in range(N)]

for i in range(M):
    u, v = [int(_) for _ in sys.stdin.readline().split()]
    edges[u-1].add(v-1)
    edges[v-1].add(u-1)


def bfs_cycle(edges, node):
    visited = [False] * N
    Q = deque()
    Q.append(node)
    parent = {}
    parent[node] = None

    while Q:
        el = Q.popleft()
        visited[el] = True
        for neighbour in edges[el]:
            if visited[neighbour]:
                if parent[el] != neighbour:
                    # print('trying cycle', el, neighbour)
                    # print(parent)
                    # backtrack
                    p1 = neighbour
                    p2 = el
                    p = {}
                    p[p1] = (0, None)
                    p[p2] = (0, None)
                    for i in range(K):
                        pp1 = parent[p1]
                        # print(pp1)
                        if pp1 is not None:
                            if pp1 in p:
                                # print('backtracked to pp1', pp1)
                                # print('1cycle len', 2 + p[pp1][0] + p[p1][0])
                                answer1 = []
                                sp = p1
                                while sp is not None:
                                    answer1.append(sp)
                                    sp = p[sp][1]
                                answer2 = []
                                sp = pp1
                                while sp is not None:
                                    answer2.append(sp)
                                    sp = p[sp][1]
                                # print(answer1, answer2)
                                return answer1 + answer2[::-1]
                            p[pp1] = (p[p1][0] + 1, p1)
                            p1 = pp1
                        pp2 = parent[p2]
                        # print(pp2)
                        if pp2 is not None:
                            if pp2 in p:
                                # print('backtracked to pp2', pp2)
                                # print('2cycle len', 2 + p[pp2][0] + p[p2][0])
                                answer1 = []
                                sp = p2
                                while sp is not None:
                                    answer1.append(sp)
                                    sp = p[sp][1]
                                answer2 = []
                                sp = pp2
                                while sp is not None:
                                    answer2.append(sp)
                                    sp = p[sp][1]
                                # print(answer1, answer2)
                                return answer1 + answer2[::-1]
                            p[pp2] = (p[p2][0] + 1, p2)
                            p2 = pp2
                continue
            visited[neighbour] = True
            parent[neighbour] = el
            Q.append(neighbour)

    return None


def bfs_set(edges, node):
    # print('from', node)
    visited = [False] * N
    Q = deque()
    Q.append(node)
    D = [-1] * N
    D[node] = 0
    parent = {}
    parent[node] = None
    answerP = [node]
    if len(answerP) >= math.ceil(K/2):
        return answerP
    answerI = []

    while Q:
        el = Q.popleft()
        visited[el] = True
        for neighbour in edges[el]:
            if visited[neighbour]:
                continue
            visited[neighbour] = True
            D[neighbour] = D[el] + 1
            parent[neighbour] = el
            if D[neighbour] % 2:
                answerI.append(neighbour)
            else:
                answerP.append(neighbour)
            if len(answerI) >= math.ceil(K/2):
                return answerI
            if len(answerP) >= math.ceil(K/2):
                return answerP
            Q.append(neighbour)

    # print(D)
    assert False


cycle = bfs_cycle(edges, 0)
if cycle:
    if len(cycle) <= K:
        print(2)
        print(len(cycle))
        print(' '.join([str(i+1) for i in cycle]))
    else:
        print(1)
        answer = []
        for i in range(math.ceil(K/2)):
            answer.append(cycle[2*i])
        print(' '.join([str(i+1) for i in answer]))
else:
    answer = bfs_set(edges, 0)
    print(1)
    print(' '.join([str(i+1) for i in answer]))
