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<int> graph[200005];
bool to[200005];
int rec[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, a, b, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  to[1] = true;
  vector<int> vertex;
  vertex.push_back(1);
  int ini, fin;
  bool flag = false;
  rec[1] = 0;
  for (int i = 0; i < n; i++) {
    int v = vertex[i];
    for (int j = 0; j < graph[v].size(); j++) {
      int u = graph[v][j];
      if (to[u]) {
        if (i - rec[u] + 2 > k) {
          ini = rec[u];
          fin = i + 1;
          flag = true;
          break;
        }
      }
    }
    if (flag) break;
    for (int j = 0; j < graph[v].size(); j++) {
      int u = graph[v][j];
      if (!to[u]) {
        to[u] = true;
        vertex.push_back(u);
        rec[v] = i + 1;
        break;
      }
    }
  }
  cout << vertex.size() - ini + 1 << endl;
  for (int i = ini - 1; i < fin; i++) cout << vertex[i] << " ";
  return 0;
}
