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>
inline int getInt() {
  int s;
  scanf("%d", &s);
  return s;
}
using namespace std;
vector<int> g[100000];
int flag[100000];
int main() {
  const int n = getInt();
  const int m = getInt();
  const int k = getInt();
  for (int i = 0; i < (int)(m); i++) {
    const int a = getInt() - 1;
    const int b = getInt() - 1;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vector<int> ans;
  vector<int> r(n);
  vector<int> f(n);
  ans.push_back(0);
  f[0] = 1;
  int idx = 0;
  for (int j = 0;; j++) {
    bool ok = false;
    for (int i : g[idx]) {
      if (f[i] == 0) {
        f[i] = 1;
        idx = i;
        ans.push_back(i);
        r[i] = j + 1;
        ok = true;
        break;
      }
    }
    if (!ok) {
      int mn = n;
      for (int k : g[idx]) {
        mn = min(mn, r[k]);
      }
      printf("%d\n", ans.size() - mn);
      for (int k = mn; k < (int)ans.size(); k++) {
        printf("%d%c", ans[k] + 1, k == (int)ans.size() - 1 ? '\n' : ' ');
      }
      break;
    }
  }
  return 0;
}
