You've got a undirected graph G, consisting of n nodes. We will consider the nodes of the graph indexed by integers from 1 to n. We know that each node of graph G is connected by edges with at least k other nodes of this graph. Your task is to find in the given graph a simple cycle of length of at least k + 1.

A simple cycle of length d (d > 1) in graph G is a sequence of distinct graph nodes v1, v2, ..., vd such, that nodes v1 and vd are connected by an edge of the graph, also for any integer i (1 ≤ i < d) nodes vi and vi + 1 are connected by an edge of the graph.

Input

The first line contains three integers n, m, k (3 ≤ n, m ≤ 105; 2 ≤ k ≤ n - 1) — the number of the nodes of the graph, the number of the graph's edges and the lower limit on the degree of the graph node. Next m lines contain pairs of integers. The i-th line contains integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the indexes of the graph nodes that are connected by the i-th edge. 

It is guaranteed that the given graph doesn't contain any multiple edges or self-loops. It is guaranteed that each node of the graph is connected by the edges with at least k other nodes of the graph.

Output

In the first line print integer r (r ≥ k + 1) — the length of the found cycle. In the next line print r distinct integers v1, v2, ..., vr (1 ≤ vi ≤ n) — the found simple cycle.

It is guaranteed that the answer exists. If there are multiple correct answers, you are allowed to print any of them.

Examples

Input

3 3 2
1 2
2 3
3 1


Output

3
1 2 3 

Input

4 6 3
4 3
1 2
1 3
1 4
2 3
2 4


Output

4
3 4 1 2 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.StringTokenizer;


public class  D {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens())
				tk=new StringTokenizer(br.readLine());
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
		public double nextDouble() throws NumberFormatException, IOException{
			return Double.valueOf(next());
		}
	}
	
	static class Node{
		int id;
		LinkedList<Integer> ve;
		int visited;
		public Node(int idd){
			id=idd;
			visited=-1;
			ve=new LinkedList<Integer>();
		}
	}
	
	static int N,M,K;
	static Node[] array;
	
	public static void main(String args[]) throws NumberFormatException, IOException{
		Scanner sc=new Scanner();
		N=sc.nextInt();
		M=sc.nextInt();
		K=sc.nextInt();
		array=new Node[N];
		for(int i=0;i<N;i++)
			array[i]=new Node(i);
		for(int i=0;i<M;i++){
			int a=sc.nextInt();
			int b=sc.nextInt();
			a--;
			b--;
			array[a].ve.add(b);
			array[b].ve.add(a);
		}
		ArrayList<Integer> chain=new ArrayList<Integer>();
		array[0].visited=0;
		chain.add(0);
		int current=0;
		int n=1;
		while(true){
			boolean available=false;
			int siguiente=0;
			int pos=Integer.MAX_VALUE;
			for(int v: array[current].ve){
				if (array[v].visited==-1){
					available=true;
					siguiente=v;
					break;
				}
				if (array[v].visited<pos){
					pos=array[v].visited;
					siguiente=v;
				}
			}
			if (available){
				chain.add(siguiente);
				array[siguiente].visited=n;
				current=siguiente;
			}
			else{
				System.out.println(n - pos);
				for(int i=pos;i<n;i++)
					System.out.print((array[chain.get(i)].id + 1)+" ");
				return;
			}
			n++;
		}
	}

}
