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;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
const int MAXN = 5 + 100000;
vector<int> g[MAXN], s;
bitset<MAXN> vis;
int k, pos[MAXN];
void dfs(int u) {
  assert(int(g[u].size()) >= k);
  vis[u] = true;
  pos[u] = s.size();
  s.push_back(u);
  int minp = 1e6;
  for (const int& v : g[u]) {
    if (!vis[v]) {
      dfs(v);
      return;
    }
    minp = min(minp, pos[v]);
  }
  cout << int(s.size()) - minp << '\n';
  for (int i = minp; i < (int)s.size(); ++i) {
    cout << 1 + s[i] << ' ';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, m;
  cin >> n >> m >> k;
  while (m--) {
    int u, v;
    cin >> u >> v;
    g[--u].push_back(--v);
    g[v].push_back(u);
  }
  vis.reset();
  s.reserve(1 + k);
  dfs(0);
  return 0;
}
