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;
int visit[100010], ans[100010], k, p;
struct edge {
  int vid, next;
};
class graph {
 private:
  int p[100010], eid;
  edge e[100010 << 1];

 public:
  graph() {
    eid = 0;
    memset(p, -1, sizeof(p));
  }
  void insert(int from, int to) {
    e[eid].vid = to;
    e[eid].next = p[from];
    p[from] = eid++;
  }
  int head(int v) { return p[v]; }
  int next(int i) { return e[i].next; }
  int getvid(int i) { return e[i].vid; }
} G;
void output() {
  printf("%d\n", p);
  for (int i = 0; i < p; i++) printf("%s%d", i ? " " : "", ans[i]);
  printf("\n");
  exit(0);
}
void dfs(int cur, int step) {
  int v;
  ans[step - 1] = cur;
  if (step > k) {
    for (int i = G.head(cur); i != -1; i = G.next(i)) {
      v = G.getvid(i);
      if (ans[0] == v) {
        p = step;
        output();
      }
    }
  }
  for (int i = G.head(cur); i != -1; i = G.next(i)) {
    v = G.getvid(i);
    if (!visit[v]) {
      visit[v] = 1;
      dfs(v, step + 1);
      visit[v] = 0;
    }
  }
}
int main() {
  int n, m, x, y;
  scanf("%d%d%d", &n, &m, &k);
  while (m--) {
    scanf("%d%d", &x, &y);
    G.insert(x, y);
    G.insert(y, x);
  }
  for (int i = 1; i <= n; i++) {
    visit[i] = 1;
    dfs(i, 1);
    visit[i] = 0;
  }
  return 0;
}
