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 
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF263D extends PrintWriter {
	CF263D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF263D o = new CF263D(); o.main(); o.flush();
	}

	int[] next, jj;
	int l_ = 1;
	int link(int l, int j) {
		next[l_] = l;
		jj[l_] = j;
		return l_++;
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		next = new int[1 + m * 2];
		jj = new int[1 + m * 2];
		int[] ao = new int[n];
		for (int h = 0; h < m; h++) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			ao[i] = link(ao[i], j);
			ao[j] = link(ao[j], i);
		}
		boolean[] used = new boolean[n];
		boolean[] used_ = new boolean[n];
		int[] qq = new int[n];
		used[0] = used_[0] = true;
		for (int i = 1; i <= n; i++) {
			int u = qq[i - 1];
			for (int l = ao[u]; l != 0; l = next[l]) {
				int v = jj[l];
				if (!used_[v]) {
					if (used[v]) {
						int h = 0;
						while (qq[h] != v)
							h++;
						println(i - h);
						while (h < i)
							print(qq[h++] + 1 + " ");
						println();
						return;
					}
					used[v] = used_[v] = true;
					qq[i] = v;
					break;
				}
			}
			if (i >= k)
				used_[qq[i - k]] = false;
		}
	}
}
