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 inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 6e5 + 10;
const double esp = 1e-9;
int cnt, nxt[maxn], head[maxn], pnt[maxn];
void add_edge(int u, int v) {
  pnt[cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt++;
}
bool used[maxn], done[maxn], flag = false;
int dis[maxn];
vector<int> path;
int n, m, k;
void dfs(int u, int deep) {
  if (used[u]) return;
  used[u] = true;
  path.push_back(u);
  dis[u] = deep;
  for (int i = head[u]; ~i && !flag; i = nxt[i]) {
    int v = pnt[i];
    if (used[v] && !done[v] && deep - dis[v] >= k) {
      int p = 0;
      while (path[p] != v) p++;
      printf("%d\n", deep - p + 1);
      for (int k = p; k <= deep; k++) {
        printf("%d ", path[k]);
      }
      printf("\n");
      flag = true;
    } else
      dfs(v, deep + 1);
  }
  done[u] = true;
  path.pop_back();
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d%d", &n, &m, &k);
  cnt = 0;
  flag = 0;
  memset(head, -1, sizeof head);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
  }
  dfs(1, 0);
  return 0;
}
