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;
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<int>> g(n);
  vector<int> p(n, -1);
  stack<int> S;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  p[0] = 0;
  for (int v = 0, w;; v = w) {
    bool flag = false;
    S.push(v);
    w = -1;
    for (int u : g[v]) {
      if (p[u] != -1 && p[v] - p[u] >= k) {
        w = u;
        flag = true;
        break;
      } else if (p[u] == -1)
        w = u;
    }
    if (!flag) {
      p[w] = 1 + p[v];
      continue;
    }
    cout << p[v] - p[w] + 1 << '\n';
    for (; S.top() != w; S.pop()) cout << S.top() + 1 << ' ';
    cout << 1 + w << '\n';
    break;
  }
}
signed main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
