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:
#!/usr/bin/env pypy3

from types import GeneratorType
from sys import stdin, stdout
import math

def input(): 
    return stdin.readline().strip()

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

class Graph:

    WHITE = 0
    GREY = 1
    BLACK = 2
    
    def __init__(self, N):
        self.N = N
        self.colours = dict()
        self.neighbours = dict()

        for v in range(1, N+1):
            self.neighbours[v] = set()
            self.colours[v] = Graph.WHITE

    @classmethod
    def line(self, N):
        g = Graph(N)
        for i in range(1, N):
            g.add_edge(i,i+1)
        return g

    def add_edge(self, u, v):
        self.neighbours[u].add(v)
        self.neighbours[v].add(u)

    def uncontractable_cycle(self):
        self.levels = dict()
        self.parent_of = dict()

        for v in range(1, self.N+1):
            if self.colours[v] == Graph.WHITE:
                r = self._uncontractable_cycle(v, v, 0)
                if r is not None: return r

        return None
        
    @bootstrap
    def _uncontractable_cycle(self, parent, u, level):
        self.levels[u] = level
        self.colours[u] = Graph.GREY
        self.parent_of[u] = parent

        back_edges = []
        
        for v in self.neighbours[u]:
            if v == parent: continue
            if self.colours[v] == Graph.GREY:
                back_edges += [(self.levels[u] - self.levels[v], (u, v))]

        if len(back_edges):
            (_, (u, v)) = sorted(back_edges)[0]
            cycle = [u]

            while True:
                if cycle[-1] == v: break
                cycle += [self.parent_of[cycle[-1]]]

            yield cycle
            return 

        for v in self.neighbours[u]:
            if v == parent: continue
            if self.colours[v] == Graph.WHITE:
                r = yield self._uncontractable_cycle(u, v, level+1)
                if r is not None:
                    yield r
            else:
                continue

        self.colours[u] = Graph.BLACK

        yield None

def ans(g, N, K):
    cycle = g.uncontractable_cycle()

    if cycle is not None:           
        if len(cycle) <= K:
            cycle = cycle[::-1]
            # print the cycle
            print(2)
            print(len(cycle))
            print(' '.join(map(str, cycle)))
        else:
            # print an independent set on the cycle
            IS = []
            for i, e in enumerate(cycle):
                if i % 2 == 1:
                    IS += [e]
            assert(len(IS) >= math.ceil(K / 2))
            IS = IS[0:math.ceil(K / 2)]
            print(1)
            print(' '.join(map(str, IS)))

        return

    
    is0 = []
    is1 = []

    for i in range(1, N+1):
        if g.levels[i] % 2 == 0:
            is0 += [i]
        else:
            is1 += [i]

    if len(is0) >= math.ceil(K / 2):
        IS = is0
    else:
        # assert(len(is1) >= math.ceil(K / 2))
        IS = is1

    IS = IS[0:math.ceil(K / 2)]
    print(1)
    print(' '.join(map(str, IS)))

N, M, K = input().split(' ')
N = int(N)
M = int(M)
K = int(K)

g = Graph(N)

for _ in range(M):
    u, v = input().split(' ')
    u = int(u)
    v = int(v)
    g.add_edge(u, v)

ans(g, N, K)