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;
vector<int> adj[100005];
int status[100005];
int dfs_number[100005];
vector<int> ans;
int done = 0;
void dfs(int now, stack<int> *s) {
  if (done) return;
  (*s).push(now);
  status[now] = 1;
  int ss = adj[now].size();
  for (int i = 0; i < ss && !done; i++) {
    int v = adj[now][i];
    if (status[v] == 0) {
      dfs_number[v] = dfs_number[now] + 1;
      dfs(v, s);
    } else if (status[v] == 1) {
      if (dfs_number[now] - dfs_number[v] >= K && !done) {
        while ((*s).top() != v) {
          ans.push_back((*s).top());
          (*s).pop();
        }
        ans.push_back(v);
        done = 1;
        return;
      }
    }
  }
  if (done) return;
  (*s).pop();
  status[now] = 2;
}
int main() {
  int a, b;
  scanf("%d %d %d", &N, &M, &K);
  for (int i = 0; i < M; i++) {
    scanf("%d %d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  stack<int> s;
  dfs_number[1] = 1;
  dfs(1, &s);
  printf("%d\n", ans.size());
  printf("%d", ans[0]);
  for (int i = 1; i < ans.size(); i++) printf(" %d", ans[i]);
  printf("\n");
  return 0;
}
