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 implements Runnable {

	int k;
	ArrayList<Integer>[] g;
	int[] col, p;

	private void Solution() throws IOException {
		int n = nextInt(), m = nextInt();
		k = nextInt();
		g = new ArrayList[n];
		for (int i = 0; i < n; i++)
			g[i] = new ArrayList<Integer>(0);
		for (int i = 0; i < m; i++) {
			int x = nextInt() - 1, y = nextInt() - 1;
			g[x].add(y);
			g[y].add(x);
		}
		Random rnd = new Random();
		while (true) {
			col = new int[n];
			p = new int[n];
			Arrays.fill(p, -1);
			int start = rnd.nextInt(n);
			dfs(start, -1);
		}
	}

	void dfs(int v, int parent) {
		col[v] = 1;
		for (int to : g[v]) {
			if (col[to] == 1 && to != parent) {
				int len = 1, cur = v;
				while (true) {
					len++;
					cur = p[cur];
					if (cur == to || cur == -1)
						break;
				}
				if (len > k) {
					println(len);
					print(to + 1);
					cur = v;
					while (true) {
						print(' ');
						print(cur + 1);
						cur = p[cur];
						if (cur == to || cur == -1)
							break;
					}
					halt();
				}
			} else if (col[to] == 0) {
				p[to] = v;
				dfs(to, v);
			}
		}
		col[v] = 2;
	}

	public static void main(String[] args) {
		new D().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer tokenizer;

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			Solution();
			out.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(261);
		}
	}

	void halt() {
		out.close();
		System.exit(0);
	}

	void print(Object... obj) {
		for (int i = 0; i < obj.length; i++) {
			if (i != 0)
				out.print(" ");
			out.print(obj[i]);
		}
	}

	void println(Object... obj) {
		print(obj);
		out.println();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}

	String nextLine() throws IOException {
		return in.readLine();
	}

	String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(nextLine());
		return tokenizer.nextToken();
	}
}
