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 n, m, k, ans[500010], l, r;
vector<int> e[500010];
map<int, bool> g[500010];
int visit[500010];
bool dfs(int u, int cnt) {
  int i, v, j;
  for (i = 0; i < k; i++) {
    v = e[u][i];
    if (!visit[v]) {
      visit[v] = cnt + 1;
      ans[cnt] = v;
      if (dfs(v, cnt + 1))
        return 1;
      else
        return 0;
    } else if (cnt - visit[v] >= k) {
      l = v;
      r = cnt;
      return 1;
    }
  }
  return 0;
}
int main() {
  int i, u, v;
  while (scanf("%d%d%d", &n, &m, &k) != EOF) {
    memset(ans, 0, sizeof(ans));
    memset(visit, 0, sizeof(visit));
    for (i = 1; i <= n; i++) e[i].clear(), g[i].clear();
    while (m--) {
      scanf("%d%d", &u, &v);
      e[u].push_back(v), e[v].push_back(u);
      g[u][v] = g[v][u] = 1;
    }
    for (i = 1; i <= n; i++) {
      if (!visit[i]) {
        ans[0] = i;
        visit[i] = 1;
        if (dfs(i, 1)) break;
      }
    }
    for (i = 0; i < n; i++)
      if (ans[i] == l) {
        l = i;
        break;
      }
    printf("%d\n", r - l);
    printf("%d", ans[l]);
    for (i = l + 1; i < r; i++) printf(" %d", ans[i]);
    puts("");
  }
}
