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;
const int maxn = 100005;
struct Edge {
  int t, next;
} edge[maxn << 1];
int head[maxn], tot;
int d[maxn], pre[maxn], vis[maxn];
int n, m, k, star;
void new_edge(int a, int b) {
  edge[tot].t = b;
  edge[tot].next = head[a];
  head[a] = tot++;
}
void out_put(int j) {
  if (j == star) {
    printf("%d ", star);
    return;
  }
  out_put(pre[j]);
  printf("%d ", j);
}
int dfs(int u, int fa, int dep) {
  if (vis[u]) {
    if (dep - d[u] >= k + 1) {
      printf("%d\n", dep - d[u]);
      star = u;
      out_put(fa);
      return 1;
    }
    return 0;
  }
  vis[u] = 1;
  d[u] = dep;
  pre[u] = fa;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int e = edge[i].t;
    if (e == fa) continue;
    if (dfs(e, u, dep + 1)) return 1;
  }
  puts("");
  vis[u] = 0;
  return 0;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  int a, b;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    new_edge(a, b);
    new_edge(b, a);
  }
  dfs(1, 0, 1);
  return 0;
}
