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>
int dfn[100010], n, m, k;
std::vector<int> tab[100010];
int stk[100010];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    tab[a].push_back(b);
    tab[b].push_back(a);
  }
  int top = 1;
  stk[top] = 1;
  while (true) {
    int now = stk[top];
    dfn[now] = top;
    int mind = n + 1, minp = -1;
    for (unsigned i = 0; i < tab[now].size(); i++) {
      int nxt = tab[now][i];
      if (dfn[nxt] < mind) {
        mind = dfn[nxt];
        minp = nxt;
      }
    }
    if (mind == 0)
      stk[top + 1] = minp;
    else {
      printf("%d\n", top - mind + 1);
      for (int i = mind; i <= top; i++)
        printf("%d%c", stk[i], i == top ? '\n' : ' ');
      return 0;
    }
    top++;
  }
}
