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 

//package D;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;
import sun.util.locale.StringTokenIterator;

public class Solution {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;
    int n, m, k, an, ff;
    int[] to, next, first, pos, a;

    public String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }
    
    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    
    void addEdge(int u, int v) {
        to[ff] = v; next[ff] = first[u]; first[u] = ff++;
        to[ff] = u; next[ff] = first[v]; first[v] = ff++;
    }
    
    void print(int l, int r) {
        System.out.println(r - l + 1);
        for (int i = l; i <= r; ++i) {
            if (i > l) System.out.print(' ');
            System.out.print(a[i] + 1);
        }
        System.out.println();
    }
    
    void run() throws Exception {
        n = nextInt();
        m = nextInt();
        k = nextInt();
        to = new int[2 * m];
        next = new int[2 * m];
        first = new int[n];
        pos = new int[n];
        a = new int[n];
        Arrays.fill(first, -1);
        Arrays.fill(pos, -1);
        for (int i = 0; i < m; ++i) {
            int u = nextInt() - 1;
            int v = nextInt() - 1;
            addEdge(u, v);
        }
        
        for (int u = 0; u != -1; ) {
            pos[u] = an;
            //System.out.println((u + 1) + " " + (an + 1));
            a[an++] = u;
            int v = -1;
            for (int i = first[u]; i != -1; i = next[i]) {
                int w = to[i];
                if (pos[w] == -1) {
                    v = w;
                    break;
                }
            }
            u = v;
        }
        boolean found = false;
        for (int i = 0; i < an; ++i) {
            int u = a[i];
            for (int j = first[u]; j != -1; j = next[j]) {
                int v = to[j];
                if (pos[v] > i && pos[v] - i >= k) {
                    print(i, pos[v]);
                    found = true;
                    break;
                }
            }
            if (found) break;
        }
    }
    
            
    public static void main(String args[]) throws Exception {
        new Solution().run();
    }
}
