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.util.*;
import java.io.*;

public class Solution{
	 static ArrayList<Integer> list = new ArrayList<>();
	 
   public static void main(String[] args)throws Exception{
	   BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	   PrintWriter out = new PrintWriter(System.out);
	   StringTokenizer st = new StringTokenizer(br.readLine());
	   
       int n = Integer.parseInt(st.nextToken()), m = Integer.parseInt(st.nextToken()), k = Integer.parseInt(st.nextToken());
		int[] from = new int[m], to = new int[m];
		for (int i = 0; i < m; i++) {
			st = new StringTokenizer(br.readLine());
			from[i] = Integer.parseInt(st.nextToken())- 1;
			to[i] = Integer.parseInt(st.nextToken()) - 1;
		}
		int[][] g = packU(n, from, to);
		
		boolean[] vis = new boolean[n];
		dfs(g, 0, vis);
		
		int ind = -1;
		for(int i = 0;ind == -1 && i < list.size(); i++) {
			for(int v : g[list.get(list.size() - 1)]) {
				if(v == list.get(i)) {
					ind = i;
					break;
				}
			}
		}
		
		out.println(list.size() - ind);
		for(int i = ind; i < list.size(); i++) {
			out.print(list.get(i) + 1 + " ");
		}
		out.close();
   }
	static void dfs(int[][] g, int u, boolean[] vis) {
		if (vis[u]) {
			return;
		}
		vis[u] = true;
 
		list.add(u);
		for (int v : g[u]) {
			if (v != 0 && !vis[v]) {
				dfs(g, v, vis);
				return;
			}
		}
	}
 
	static int[][] packU(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
			g[to[i]][--p[to[i]]] = from[i];
		}
		return g;
	}
}