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>
const int MAX = 100014;
using namespace std;
int nodgasit, taty, tata[MAX], dist[MAX], k, n;
vector<int> gr[MAX];
void dfs(int nod) {
  if (nodgasit) return;
  for (vector<int>::iterator it = gr[nod].begin(); it != gr[nod].end(); ++it)
    if (tata[nod] != (*it)) {
      if (!tata[*it]) {
        tata[*it] = nod;
        dist[*it] = dist[nod] + 1;
        dfs(*it);
      } else {
        if (dist[nod] - dist[*it] >= k) {
          nodgasit = *it;
          taty = nod;
          return;
        }
      }
    }
}
int main() {
  int m;
  cin >> n >> m >> k;
  for (; m; --m) {
    int x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  tata[1] = -1;
  dfs(1);
  cout << dist[taty] - dist[nodgasit] + 1 << '\n';
  while (nodgasit != taty) {
    cout << taty << ' ';
    taty = tata[taty];
  }
  cout << taty << '\n';
  return 0;
}
