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 long long mod = 1e9 + 7;
const long long maxn = 1e5 + 7;
const long long maxm = 1e7 + 7;
int n, m, k;
struct EDGE {
  int to, next;
} p[maxn * 2];
int head[maxn], cnt = 0;
void add(int u, int v) {
  p[cnt].to = v;
  p[cnt].next = head[u];
  head[u] = cnt++;
}
int deep[maxn * 2], ans[maxn * 2], d = 0;
bool endf = false;
void dfs(int u) {
  if (endf) return;
  deep[u] = ++d;
  ans[d] = u;
  for (register int i = head[u]; i != -1; i = p[i].next) {
    if (endf) return;
    int v = p[i].to;
    if (deep[v]) {
      if (deep[u] - deep[v] >= k) {
        printf("%d", deep[u] - deep[v] + 1);
        printf("\n");
        ;
        for (register int i = deep[v]; i <= deep[u]; i = i + 1) {
          printf("%d", ans[i]);
          printf("%s", " ");
        }
        endf = true;
        break;
      }
    } else
      dfs(v);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  int u, v;
  for (register int i = 1; i <= m; i = i + 1) {
    scanf("%d", &u);
    scanf("%d", &v);
    add(u, v);
    add(v, u);
  }
  dfs(1);
}
