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;
int col[100009], n, m, k, d;
vector<int> adj[100009];
vector<int> res;
bool dfs(int u, int dep);
int main() {
  int a, b;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int rescnt = 0;
  for (int i = 1; i <= n; i++) {
    col[i] = 1;
    dfs(i, 1);
    if (((int)res.size())) break;
    col[i] = 0;
  }
  printf("%d\n", ((int)res.size()));
  for (int i = ((int)res.size()) - 1; i >= 0; i--) printf("%d ", res[i]);
  puts("");
  return 0;
}
bool dfs(int u, int dep) {
  for (int i = 0; i < ((int)adj[u].size()); i++) {
    int v = adj[u][i];
    if (!col[v]) {
      col[v] = dep + 1;
      if (dfs(v, dep + 1)) {
        if (d <= dep) res.push_back(u);
        return true;
      }
      col[v] = 0;
    } else {
      if (col[v] <= dep - k) {
        d = col[v];
        res.push_back(u);
        return true;
      }
    }
  }
  return false;
}
