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.*;

public class Main
{
    static Vector<Vector<Integer>> gr;
    static int[] color, time_in;
    static int dfs_timer = 0;
    static boolean f = true;
    static int len, res_size;
    static Vector<Integer> r;

    static void dfs (int v)
    {
        time_in[v] = dfs_timer++;
        color[v] = 1;
        for (int i = 0; i < gr.get(v).size() && f; ++i)
        {
            int to = gr.get(v).get(i);
            if (color[to] == 0)
                dfs(to);
            else if (color[to] == 1 && time_in[v] - time_in[to] + 1 > len)
            {
                res_size = time_in[v] - time_in[to] + 1;
                f = false;
            }
        }
        r.add(v);
        color[v] = 2;
    }

    public static void main(String[] args)
    {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt(), m = cin.nextInt(), k = cin.nextInt();
        len = k;
        gr = new Vector<Vector<Integer>>();
        r = new Vector<Integer>();
        gr.setSize(n);
        color = new int[n];
        time_in = new int[n];
        for (int i = 0; i < n; ++i)
        {
            gr.set(i, new Vector<Integer>());
        }
        for (int i = 0; i < m; ++i)
        {
            Integer a = cin.nextInt()-1, b = cin.nextInt()-1;
            gr.get(a).add(b);
            gr.get(b).add(a);
        }
        for (int i = 0; i < n; ++i)
        {
            if (color[i] == 0 && f == true)
                dfs(i);
        }
        System.out.println(res_size);
        for (int i = 0; i < r.size() && i < res_size; ++i)
            System.out.print(r.get(i)+1 + " ");
    }
}
