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;
vector<vector<int>> G;
vector<bool> vis;
vector<int> h, p;
int n, m, k;
void DFS(int u) {
  vis[u] = true;
  for (auto &v : G[u]) {
    if (!vis[v]) {
      vis[v] = true;
      p[v] = u;
      h[v] = h[u] + 1;
      DFS(v);
    } else if (h[u] - h[v] + 1 > k) {
      int x = u;
      cout << h[u] - h[v] + 1 << endl;
      for (int i = 0; i < h[u] - h[v] + 1; i++) {
        cout << x + 1 << " ";
        x = p[x];
      }
      exit(0);
    }
  }
}
int main() {
  int u, v;
  cin >> n >> m >> k;
  G.resize(n);
  vis.resize(n);
  fill_n(vis.begin(), n, false);
  h.resize(n);
  p.resize(n);
  while (m--) {
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  DFS(0);
  return 0;
}
