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;
set<pair<int, int> > s;
std::vector<int> vv;
std::vector<std::vector<int> > gr;
std::vector<int> vis;
int k;
pair<int, int> dfs(int u, int l) {
  vis[u] = l;
  vv.emplace_back(u);
  for (int v : gr[u]) {
    if ((vis[v]) && (vis[u] - vis[v]) >= k) {
      return {vis[v], vis[u]};
    } else if (!vis[v]) {
      return dfs(v, l + 1);
    }
  }
  return {-1, -1};
}
int32_t main() {
  int n;
  cin >> n;
  int m;
  cin >> m;
  cin >> k;
  gr.resize(n);
  for (int i = 0; i < m; ++i) {
    int u;
    cin >> u;
    u--;
    int v;
    cin >> v;
    v--;
    gr[u].emplace_back(v);
    gr[v].emplace_back(u);
  }
  vis.resize(n, 0);
  auto pp = dfs(0, 1);
  cout << pp.second - pp.first + 1 << "\n";
  for (int i = pp.first - 1; i < pp.second; ++i) {
    cout << vv[i] + 1 << " ";
  }
}
