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;
long long int st[100005] = {0}, vis[100005] = {0}, pr[100005] = {0};
vector<long long int> g[100005];
long long int mx, lst, k;
bool dfs(long long int x, long long int par, long long int t) {
  if (vis[x]) {
    if (t - st[x] > k) {
      mx = t - st[x];
      lst = par;
      return 1;
    } else
      return 0;
  }
  vis[x] = 1;
  st[x] = t;
  pr[x] = par;
  long long int f = 0;
  for (long long int i = 0; i < g[x].size(); i++) {
    if (g[x][i] == par) continue;
    f |= dfs(g[x][i], x, t + 1);
    if (f) return f;
  }
  if (f == 0) vis[x] = 0;
  return f;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, x, y;
  cin >> n >> m >> k;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  bool r = dfs(1, 1, 1);
  cout << mx << endl;
  while (mx--) {
    cout << lst << " ";
    lst = pr[lst];
  }
}
