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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class r649d {
	public static void main(String[] args) {
		FastScanner scan=new FastScanner();
		PrintWriter out=new PrintWriter(System.out);
		int n=scan.nextInt(), m=scan.nextInt(), k=scan.nextInt();
		a=new ArrayList[n];
		for(int i=0;i<n;i++) a[i]=new ArrayList<>();
		for(int i=0;i<m;i++) {
			int u=scan.nextInt()-1, v=scan.nextInt()-1;
			a[u].add(v);
			a[v].add(u);
		}
		if(m==n-1) {
			//tree
			depth=new int[n];
			init(0,-1,0);

			int e=0;
			for(int i=0;i<n;i++) {
				if(depth[i]%2==0) e++;
			}
			int mod;
			if(e>=(k+1)/2) mod=0;
			else mod=1;

			System.out.println(1);
			int ct=0;
			for(int i=0;i<n;i++) {
				if(depth[i]%2==mod) {
					out.print((i+1)+" ");
					ct++;
				}
				if(ct==(k+1)/2) break;
			}
		}
		else {
			/*
			we need to find any cycle such that there
			is no edge cutting through it
			we know that such a cycle will always exist
			because we already got rid of the case where
			the graph is a tree

			if this cycle has length <= k: we have answered the second query
			else we take every other node in the cycle to answer
			the first query
			 */
			pre=new int[n];
			Arrays.fill(pre,-1);
			low=new int[n];
			par=new int[n];
			cycle=new ArrayList<>();

			dfs(0,-1);

			if(cycle.size()<=k) {
				out.println(2);
				out.println(cycle.size());
				for(int i=0;i<cycle.size();i++) {
					out.print(cycle.get(i));
					out.print(" ");
				}
			}
			else {
				System.out.println(1);
				int ct=0;
				for(int i=0;i<cycle.size();i+=2) {
					out.print(cycle.get(i));
					out.print(" ");
					ct++;
					if(ct==(k+1)/2) break;
				}
			}
		}
		out.close();
	}
	static int ct=0;
	static int[] pre,low,par;
	static ArrayList<Integer> cycle;

	//return true if we found a cycle
	public static boolean dfs(int at, int prev) {
		par[at]=prev;
		pre[at]=ct++;
		int ans=-1;//ans=deepest node connected to at by a back edge

		for(int nxt:a[at]) {
			if(nxt!=prev&&pre[nxt]!=-1) {
				//we have a back edge from at to nxt, let's update the answer
				ans=Math.max(ans,nxt);
			}
		}
		if(ans!=-1) {
			//we have a cycle
			cycle.add(ans+1);
			int cur=at;
			while(true) {
				cycle.add(cur+1);
				cur=par[cur];
				if(cur==ans) break;
			}
			return true;
		}
		for(int nxt:a[at]) {
			if(pre[nxt]==-1) {
				if(dfs(nxt,at)) return true;
			}
		}
		return false;
	}
	static ArrayList<Integer>[] a;
	public static void init(int at, int prev, int d) {
		depth[at]=d;
		for(int nxt:a[at]) {
			if(nxt==prev) continue;
			init(nxt,at,d+1);
		}
	}
	static int[] depth;
	
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e){e.printStackTrace();}
		}

		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}

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

		public long nextLong() {return Long.parseLong(next());}

		public double nextDouble() {return Double.parseDouble(next());}

		public String nextLine() {
			String line = "";
			if(st.hasMoreTokens()) line = st.nextToken();
			else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
			while(st.hasMoreTokens()) line += " "+st.nextToken();
			return line;
		}
	}
}