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.util.*;
import java.io.*;
public class Main{
    static int n,m,k;
    static ArrayList<Integer> arr[];
    static StringBuffer ans=new StringBuffer();
    public static void main(String args[])throws IOException{
        BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
        String inp[]=br.readLine().trim().split(" ");
        n = Integer.parseInt(inp[0]);
        m = Integer.parseInt(inp[1]);
        k = Integer.parseInt(inp[2]);
        
        arr=new ArrayList[n+1];
        for(int i=0;i<=n;i++)   arr[i]=new ArrayList<>();
        for(int i=0;i<m;i++){
            inp=br.readLine().trim().split(" ");
            int a =Integer.parseInt(inp[0]);
            int b =Integer.parseInt(inp[1]);
            arr[a].add(b);
            arr[b].add(a);
        }
        
        boolean vis[]=new boolean[n+1];
        int color[]=new int[n+1];
        for(int i=1;i<=n;i++){
            if(!vis[i]){
               boolean cycle= dfs(i,i,vis,color);
               if(cycle){
                   System.out.print(ans);
                   System.exit(0);
               }
            }
        }
        /*
            Can't find any cycle 
            Hence its a tree
            Hence, We can take color with most frequency and print required nodes
        */
        
        //Tree Coloring
        ans.append(1+"\n");
        ArrayList<Integer> col1=new ArrayList<>();
        ArrayList<Integer> col2=new ArrayList<>();
        for(int i=1;i<=n;i++){
            if(color[i]==0) col1.add(i);
            else    col2.add(i);
        }
        
        if(col1.size()>=col2.size()){
            int i=0;
            while(k-->0){
                ans.append(col1.get(i++)+" ");
                k--;
            }
        }
        else{
            int i=0;
            while(k-->0){
                ans.append(col2.get(i++)+" ");
                k--;
            }
        }
        System.out.print(ans);
    }
    
    static boolean dfs(int node,int par,boolean vis[],int color[]){
        vis[node]=true;
        for(int i=0;i<arr[node].size();i++){
            int in = arr[node].get(i);
            if(!vis[in]){
                color[in]=color[node]^1;
                boolean temp=dfs(in,node,vis,color);
                if(temp)
                return temp;
            }
            else if(in != par){//Cycle node->x->y->z->in
                
            //Finding nodes involved in cycle--------
                Queue<Integer> q=new LinkedList<>();
                int parent[]=new int[n+1];
                boolean visited[]=new boolean[n+1];
                q.add(node);
                visited[node]=true;
                while(!q.isEmpty()){
                    int nd =q.poll();
                    for(int ii=0;ii<arr[nd].size();ii++){
                        int index=arr[nd].get(ii);
                        if(!visited[index] && !(nd==node && index == in)){
                            visited[index]=true;
                            parent[index]=nd;
                            q.add(index);
                        }
                    }
                }
                ArrayList<Integer> li=new ArrayList<>();
                li.add(node);
                int temp_node = in;
                while(temp_node != node){
                    li.add(temp_node);
                    temp_node=parent[temp_node];
                }
             
            //If cycle length <= k ; print entire cycle 
                if(li.size()<=k){
                    ans.append(2+"\n"+li.size()+"\n");
                    for(int x:li)
                    ans.append(x+" ");
                }
            //Else print alternate nodes
                else{
                    ans.append(1+"\n");
                    int req=k%2==0?k/2:k/2+1;
                    int ii=0;
                    while(req-->0){
                        ans.append(li.get(ii)+" ");
                        ii+=2;
                    }
                }
                return true;
            }
        }
        return false;
    }
    
}