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;
const int MAX_N = 1000 * 100;
int mark[MAX_N];
int n, m, k, tst = -1, o = 0;
vector<int> adj[MAX_N];
vector<int> ans;
int dfs(int v, int c) {
  mark[v] = c;
  for (int i = 0; i < adj[v].size(); i++) {
    if (mark[adj[v][i]] == 0) {
      int p = dfs(adj[v][i], c + 1);
      if (p == 1) {
        if (v == tst) {
          return 0;
        }
        ans.push_back(v + 1);
        return 1;
      } else
        return 0;
    } else if (mark[v] - mark[adj[v][i]] >= k && tst < 0) {
      ans.push_back(adj[v][i] + 1);
      ans.push_back(v + 1);
      tst = adj[v][i];
      return 1;
    }
  }
  return 0;
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cout << endl << endl;
  for (int i = 0; i < n; i++) {
    if (mark[i] == 0) {
      dfs(i, 1);
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}
