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.io.PrintWriter;
import java.lang.management.BufferPoolMXBean;
import java.math.BigInteger;
import java.util.*;

public class DFSs {

   static ArrayList<Integer>[] g;
   static int[] cl;
   static int[] p;
   static boolean[] visited;
    static int cycle_st, cycle_end,cylenum;
    static int cnt,k;
        //Cycle checkinf
    static StringBuilder f;
    private static int[] dpth;

    static boolean CycleCheck (int v) {
        //color vertex to gray
         cl[v]=1;
        for (int i = 0; i <g [v].size(); ++i) {
            int to = g [v].get(i);
            if (cl [to] == 0) {
                p[to]=v;
                dpth[to] = dpth[v]+1;
                if (CycleCheck (to)) {
                    return true;
                }
            }
            else if (to!=p[v]&&Math.abs(dpth[v]-dpth[to])>=k) {
                cycle_end = v;
                cycle_st = to;
                return true;
            }
        }
        //color vertex that is not cycle from v
        cl[v]=2;
        return false;
    }







    public static void main(String[] args) throws IOException {
        BufferedReader in= new BufferedReader(new InputStreamReader(System.in));
        try (PrintWriter or = new PrintWriter(System.out)){
            StringTokenizer b= new StringTokenizer(in.readLine());
            int n=Integer.parseInt(b.nextToken());int m=Integer.parseInt(b.nextToken());k=Integer.parseInt(b.nextToken());
                g= new ArrayList[n+1];
            for (int i = 0; i < n; i++) {
                g[i]=new ArrayList<>();
            }
           // v= new boolean[n][n];
            cl=new int[n+1];
            p= new int[n+1];
            dpth = new int[n+1];
            Arrays.fill(p,-1);
            for (int i = 0; i < m; i++) {
                b=new StringTokenizer(in.readLine());
                int x=Integer.parseInt(b.nextToken())-1;
                int y=Integer.parseInt(b.nextToken())-1;
               g[x].add(y);
               g[y].add(x);
            }

            CycleCheck(0);
                   ArrayList <Integer> cycle=new ArrayList<>();
                       for (int v = cycle_end; v!= cycle_st; v = p[v]) cycle.add(v);
                   cycle.add (cycle_st);
                       or.println(cycle.size());
                       for (int p = cycle.size()-1; p>=0; --p)
                           or.print((cycle.get(p) +1) +" ");

               }
            }


}
