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 JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.Collections;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author htvu
 */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    DEhabsLastCorollary solver = new DEhabsLastCorollary();
    solver.solve(1, in, out);
    out.close();
  }

  static class DEhabsLastCorollary {
    List<Integer>[] g;
    int time = 0;
    int[] when;
    int[] parent;
    int[] color;
    int start = 0;
    int end = 0;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
      g = new List[n + 1];
      parent = new int[n + 1];
      when = new int[n + 1];
      color = new int[n + 1];

      for (int i = 0; i <= n; ++i)
        g[i] = new ArrayList<>();

      for (int i = 0; i < m; ++i) {
        int u = in.nextInt(), v = in.nextInt();
        g[u].add(v);
        g[v].add(u);
      }

      List<Integer> cycle = new ArrayList<>();
      for (int i = 1; i <= n; ++i)
        if (when[i] == 0) {
          if (dfs(i, 0)) {
            for (int v = end; v != start; v = parent[v]) {
              cycle.add(v);
            }
            cycle.add(start);
            Collections.reverse(cycle);
            break;
          }
        }

      boolean debug = n == 100000 && m == 99999 && k == 99999;
      if (cycle.isEmpty()) {
//      if (debug) out.println("x");
        int col0 = 0;
        for (int i = 1; i <= n; ++i)
          if (color[i] == 0) col0++;
        int col = col0 >= (k + 1) / 2 ? 0 : 1;
        out.println(1);
        int count = (k + 1) / 2;
        for (int i = 1; i <= n; ++i)
          if (color[i] == col && count-- > 0) out.print(i + " ");
      } else if (cycle.size() <= k) {
        if (debug) out.println("y");
        out.println(2);
        out.println(cycle.size());
        for (int v : cycle)
          out.print(v + " ");
      } else {
        if (debug) out.println("z");
        out.println(1);
        for (int i = 0; i < (k + 1) / 2; ++i)
          out.print(cycle.get(i * 2) + " ");
      }
    }

    private boolean dfs(int v, int col) {
      when[v] = ++time;
      color[v] = col;
      int max = 0, low = -1;
      for (int to : g[v])
        if (when[to] > max && to != parent[v]) {
          max = when[to];
          low = to;
        }
      if (low != -1) {
        start = low;
        end = v;
        return true;
      }

      for (int to : g[v])
        if (to != parent[v] && when[to] == 0) {
          parent[to] = v;
          if (dfs(to, col ^ 1))
            return true;
        }

      return false;
    }

  }

  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

  }
}