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 D263 {
    public static void main(String args[]) {
        new D263().solve();
    }

    int k;
    Map<Integer, ArrayList<Integer>> adj = new HashMap<Integer,ArrayList<Integer>>();
    public void solve() {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        k = in.nextInt();
        for (int i = 1; i <= n; i++) {
            adj.put(i, new ArrayList<Integer>());
        }

        for (int i = 1; i <= m; i++) {
            int a =in.nextInt();
            int b = in.nextInt();
            adj.get(a).add(b);
            adj.get(b).add(a);
        }

        for (int i = 1; i <= n; i++) {
            int[] visited = new int[n + 1];
            start = i;
            if (dfs(i, i, visited)) {
                System.out.println(path.size());
                for (int j : path) {
                    System.out.print(j + " ");
                }
                return;
            }
        }

    }

    LinkedList<Integer> path = new LinkedList<Integer>();
    ArrayList<Integer> answer = new ArrayList<Integer>();
    int start;

    boolean dfs(int from, int now, int[] v) {
        v[now] = 1;
        path.addLast(now);
        for (int to : adj.get(now)) {
            if (to == start && start != from) {
                if (path.size() >= k + 1) {
                    answer.addAll(path);
                    return true;
                }
            } else if (v[to] == 0) {
                if (dfs(now, to, v)) {
                    return true;
                }
            }
        }
        path.removeLast();
        return false;
    }
}
