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 
#include <bits/stdc++.h>
using namespace std;
vector<int> road[100011];
int n, m, k;
int vis[100111];
int line[100111];
bool dfs(int p, int d) {
  if (d > k && line[p]) {
    printf("%d\n", d);
    return 1;
  }
  for (int i = 0; i < road[p].size(); i++) {
    if (!vis[road[p][i]]) {
      vis[road[p][i]] = 1;
      if (dfs(road[p][i], d + 1)) {
        printf("%d ", road[p][i]);
        return 1;
      }
      break;
    }
  }
  return 0;
}
int main() {
  while (scanf("%d%d%d", &n, &m, &k) > 0) {
    for (int i = 0; i < m; i++) {
      int t1, t2;
      scanf("%d%d", &t1, &t2);
      road[t1].push_back(t2);
      road[t2].push_back(t1);
    }
    memset(vis, 0, sizeof(vis));
    ;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) {
        vis[i] = 1;
        memset(line, 0, sizeof(line));
        ;
        for (int j = 0; j < road[i].size(); j++) line[road[i][j]] = 1;
        if (dfs(i, 1)) {
          printf("%d", i);
          break;
        }
      }
  }
  return 0;
}
