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 maxn = 100005;
vector<int> edge[maxn];
bool vis[maxn];
int pre[maxn], d[maxn];
int u, v, n, m, k, last, dep;
bool mark;
void dfs(int id, int cnt) {
  if (mark) return;
  for (int i = 0; i < edge[id].size(); i++) {
    if (mark) return;
    int x = edge[id][i];
    if (x == u && cnt >= k) {
      mark = true;
      dep = cnt + 1;
      last = id;
      return;
    }
    if (vis[x]) continue;
    vis[x] = true;
    pre[x] = id;
    dfs(x, cnt + 1);
    vis[x] = false;
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    vis[i] = false;
    d[i] = 0;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
    d[u]++;
    d[v]++;
  }
  u = 1;
  for (int i = 2; i <= n; i++) {
    if (d[i] > d[u]) u = i;
  }
  vis[u] = true;
  mark = false;
  dfs(u, 0);
  printf("%d\n", dep);
  int i = last;
  for (; i != u; i = pre[i]) printf("%d ", i);
  printf("%d\n", u);
  return 0;
}
