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;
const int MAXN = 1e5 + 100;
vector<int> adj[MAXN];
deque<int> ans;
bool seen[MAXN];
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  while (m--) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  ans.push_back(1);
  seen[1] = true;
  set<int> bef;
  for (int add = 2;; add++) {
    int last = ans[ans.size() - 1];
    for (int i = 0; i < adj[last].size(); ++i)
      if (bef.find(adj[last][i]) != bef.end()) {
        int begin = adj[last][i];
        while (ans[0] != begin) ans.pop_front();
        cout << ans.size() << endl;
        for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
        cout << endl;
        return 0;
      }
    for (int i = 0; i < adj[last].size(); ++i) {
      int next = adj[last][i];
      if (seen[next] == false) {
        ans.push_back(next);
        seen[next] = true;
        break;
      }
    }
    if (add - k >= 1) {
      bef.insert(ans[add - k - 1]);
    }
  }
  return 0;
}
