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.*;
import java.util.*;

public class D {
    public static void main(String[] args) throws IOException {
        new D().solve();
    }

    void solve() throws IOException {
        // Scanner sc = new Scanner(new File("input.txt"));
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] sp;

        sp = in.readLine().split(" ");

        int n = Integer.parseInt(sp[0]);
        int m = Integer.parseInt(sp[1]);
        int k = Integer.parseInt(sp[2]);

        LinkedList<Integer>[] edge = new LinkedList[n];
        for (int i = 0; i < n; i++) {
            edge[i] = new LinkedList<Integer>();
        }
        for (int i = 0; i < m; i++) {
            sp = in.readLine().split(" ");
            int a = Integer.parseInt(sp[0]) - 1;
            int b = Integer.parseInt(sp[1]) - 1;
            edge[a].add(b);
            edge[b].add(a);
        }
        ArrayList<Integer> visit = new ArrayList<Integer>(n);
        boolean[] bvisit = new boolean[n];
        boolean[] ok = new boolean[n];

        int idx = 0;
        while (true) {
            visit.add(idx);
            bvisit[idx] = true;
            if (visit.size() >= k + 1) {
                // ok?
                ok[visit.get(visit.size() - (k + 1))] = true;
                for (int next : edge[idx]) {
                    if (ok[next]) {
                        // OK!
                        // next以降を出力
                        int startIdx = visit.indexOf(Integer.valueOf(next));
                        StringBuilder str = new StringBuilder((visit.size() - startIdx) + "\n");
                        for (int i = startIdx; i < visit.size(); i++) {
                            str.append((visit.get(i) + 1) + " ");
                        }
                        System.out.println(str);
                        return;
                    }
                }
            }
            for (int next : edge[idx]) {
                if (!bvisit[next]) {
                    // まだ
                    idx = next;
                    break;
                }
            }
        }
    }
}

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