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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 100010;
struct edge {
  int v, nxt;
} e[N << 1];
int head[N], cnt, n, m, k, x, y, dep[N], fa[N], ans, out[N];
inline void add(int u, int v) {
  e[++cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
inline bool dfs(int u) {
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa[u]) continue;
    if (!fa[v]) {
      fa[v] = u;
      dep[v] = dep[u] + 1;
      if (dfs(v)) return 1;
    } else if (dep[u] - dep[v] >= k) {
      ans = dep[u] - dep[v] + 1;
      for (int j = u; j != v; j = fa[j]) {
        out[++out[0]] = j;
      }
      out[++out[0]] = v;
      return 1;
    }
  }
  return 0;
}
int main() {
  n = read();
  m = read();
  k = read();
  for (int i = 1; i <= m; ++i) {
    x = read();
    y = read();
    add(x, y);
    add(y, x);
  }
  bool csyakioi = dfs(fa[1] = 1);
  printf("%d\n", ans);
  for (int i = 1; i <= out[0]; ++i) printf("%d ", out[i]);
  return 0;
}
