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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.System.out;

public class D {

   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int cantNodos = sc.nextInt();
      int cantAristas = sc.nextInt();
      int minGrado = sc.nextInt();

      ArrayList<ArrayList<Integer>> grafo =
            new ArrayList<ArrayList<Integer>>(cantNodos);

      for (int i = 0; i < cantNodos; i++) {
         grafo.add(i, new ArrayList<Integer>());
      }

      for (int i = 0; i < cantAristas; i++) {
         int n1 = sc.nextInt() - 1;
         int n2 = sc.nextInt() - 1;
         grafo.get(n1).add(n2);
         grafo.get(n2).add(n1);
      }

      sc.close();

      ArrayList<Integer> camino = new ArrayList<Integer>();
      int[] orden = new int[cantNodos];
      Arrays.fill(orden, -1);

      camino.add(0);
      int nuevoNodo = 0;

      while (orden[nuevoNodo] == -1) {
         orden[nuevoNodo] = camino.size();
         nuevoNodo =
               getNodoQueNoEsteEnLosUltimosK(grafo, camino, orden,
                     minGrado, camino.get(camino.size() - 1));
         camino.add(nuevoNodo);
      }

      camino.subList(0, orden[nuevoNodo]).clear();
      out.println(camino.size());

      StringBuilder res = new StringBuilder();
      for (int nodo : camino) {
         res.append(nodo + 1).append(" ");
      }
      res.deleteCharAt(res.length() - 1);
      out.println(res);
   }

   private static Integer getNodoQueNoEsteEnLosUltimosK(
         ArrayList<ArrayList<Integer>> grafo, ArrayList<Integer> camino,
         int[] orden, int minGrado, int nodoActual) {
      int j = -1;
      for (Integer i : grafo.get(nodoActual)) {
         if (orden[i] == -1 || orden[i] <= camino.size() - minGrado) {
            j = i;
            break;
         }
      }
      return j;
   }
}
