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.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;


public class D {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		D solver = new D();
		solver.solve(in, out);
		in.close();
		out.close();
	}
	
	private int counter, count, minValue, length, start, end;
	private ArrayList< ArrayList<Integer> > graph;
	private int[] color, parent;
	
	private void solve(Scanner in, PrintWriter out) {
		counter = in.nextInt();
		count = in.nextInt();
		minValue = in.nextInt();
		length = 0; 
		end = 0;
		color = new int[counter];
		parent = new int[counter];
		graph = new ArrayList< ArrayList<Integer> >(counter);
		for (int i = 0; i < counter; ++i) {
			graph.add(new ArrayList<Integer>());
		}
		int a, b;
		for (int i = 0; i < count; ++i) {
			a = in.nextInt() - 1;
			b = in.nextInt() - 1;
			graph.get(a).add(b);
			graph.get(b).add(a);
		}
		for (int i = 0; i < counter; ++i) {
			start = i;
			Arrays.fill(color, 0);
			Arrays.fill(parent, -1);
			if (dfs(i)) 
				break;
		}
		out.println(length);
		int cur = end;
		for (int i = 0; i < length; ++i) {
			out.print(cur + 1);
			out.print(" ");
			cur = parent[cur];
		}	
	}
	
	private boolean dfs(int vertex) {
		color[vertex] = 1;
		++length;
		boolean good = false;
		for (int v : graph.get(vertex)) {
			if (color[v] == 0) {
				parent[v] = vertex;
				if (dfs(v))
					return true;
			} else if (color[v] == 1) 
				if (v == start) {
					if (length > minValue) {
						end = vertex;
						return true;
					} 
					good = true;
				}
		}
		if (good)
			color[vertex] = 0;
		else
			color[vertex] = 2;
		--length;
		return false;
	}

}