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.ArrayDeque;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

/*
4 4 4
1 2
2 3
3 4
3 1
 */
public class D {
	
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		PrintWriter out=new PrintWriter(System.out);
		int n=fs.nextInt(), m=fs.nextInt(), k=fs.nextInt();
		Node[] nodes=new Node[n];
		for (int i=0; i<n; i++) nodes[i]=new Node(i);
		for (int i=0; i<m; i++) {
			int a=fs.nextInt()-1,b=fs.nextInt()-1;
			nodes[a].adj.add(nodes[b]);
			nodes[b].adj.add(nodes[a]);
		}
		
		if (m==n-1) {
			//TREE
			nodes[0].twoCol(0, null);
			ArrayList<Node>[] ofCol=new ArrayList[2];
			for (int i=0; i<2; i++) ofCol[i]=new ArrayList<>();
			for (Node nn:nodes) ofCol[nn.col].add(nn);
			out.println(1);
			ArrayList<Node> ans=ofCol[0].size()>=(k+1)/2?ofCol[0]:ofCol[1];
			for (int i=0; i<(k+1)/2; i++) 
				out.print(ans.get(i)+" ");
			out.println();
			out.close();
		}
		else {
			//FIND CYCLE
			nodes[0].dfs(null, new ArrayDeque<>());
			int cycleLen=cycle.size();
			for (int i=0; i<cycle.size(); i++) cycle.get(i).indexInCycle=i;
			
			int curBest=cycleLen;
			Node curStart=cycle.get(0), curEnd=cycle.get(cycleLen-1);
			for (Node a:cycle) {
				for (Node b:a.adj) {
					if (b.indexInCycle==-1) {
						continue;
					}
					Node first=a.indexInCycle<b.indexInCycle?a:b;
					Node second=a.indexInCycle<b.indexInCycle?b:a;
					int len1=second.indexInCycle-first.indexInCycle+1;
					if (len1!=2 && len1<curBest) {
						curBest=len1;
						curStart=first;
						curEnd=second;
					}
					int len2=cycleLen-second.indexInCycle+first.indexInCycle+1;
					if (len2!=2 && len2<curBest) {
						curBest=len2;
						curStart=second;
						curEnd=first;
					}
				}
			}
			
			ArrayList<Node> ans=new ArrayList<>();
			Node at=curStart;
			while (at!=curEnd) {
				ans.add(at);
				at=cycle.get((at.indexInCycle+1)%cycleLen);
			}
			ans.add(curEnd);
			
			if (ans.size()<=k) {
				out.println(2);
				out.println(ans.size());
				for (Node nn:ans)
					out.print(nn+" ");
				out.println();
			}
			else {
				out.println(1);
				for (int i=0; i<(k+1)/2; i++) {
					out.print(ans.get(i*2)+" ");
				}
				out.println();
			}
			out.close();
		}
	}
	
	static ArrayList<Node> cycle=new ArrayList<>();
	
	static class Node {
		int index;
		ArrayList<Node> adj=new ArrayList<>();
		int col=-1;
		boolean visited=false;
		int indexInCycle=-1;
		
		public Node(int index) {
			this.index=index;
		}
		public void twoCol(int col, Node par) {
			this.col=col;
			for (Node nn:adj) {
				if (nn==par) continue;
				nn.twoCol(col^1, this);
			}
		}
		//returns true if a cycle has been found
		public boolean dfs(Node par, ArrayDeque<Node> stk) {
			if (visited) {
				//pop off the stack until we get to me
				while (stk.peekLast()!=this) {
					cycle.add(stk.removeLast());
				}
				cycle.add(this);
				return true;
			}
			else {
				visited=true;
				stk.addLast(this);
				for (Node nn:adj) {
					if (nn==par) continue;
					if (nn.dfs(this, stk)) {
						return true;
					}
				}
				if (stk.removeLast() != this) throw null;
				return false;
			}
		}
		
		public String toString() {
			return ""+(index+1);
		}
	}

	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

	
}