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> graph[1000000];
int recursionS[1000000];
int p[1000000], found = 0, st, n, m, k, l;
void dfs(int x, int pos, int par) {
  int u;
  recursionS[x] = pos;
  for (int i = 0; i < graph[x].size() && found == 0; i++) {
    u = graph[x][i];
    if (recursionS[u] && u != par) {
      if (recursionS[x] - recursionS[u] >= k)
        found = u, st = x, l = recursionS[x] - recursionS[u];
      else
        continue;
    } else if (!recursionS[u])
      p[u] = x, dfs(u, pos + 1, x);
  }
  recursionS[x] = 0;
  return;
}
int main() {
  cin >> n >> m >> k;
  int u, v;
  for (int i = 0; i < m; i++)
    cin >> u >> v, graph[u].push_back(v), graph[v].push_back(u);
  dfs(1, 1, -1);
  cout << l + 1 << endl;
  cout << st << " ";
  while (st != found) st = p[st], cout << st << " ";
  return 0;
}
