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;
vector<int> g[100005];
bool visit[100005];
int parr[100005];
int dfs(int s, vector<int> &path) {
  visit[s] = 1;
  path.push_back(s);
  bool flag = 0;
  for (int v = 0; v < g[s].size(); v++) {
    if (!visit[g[s][v]]) {
      flag = 1;
      int d = dfs(g[s][v], path);
      if (d != -1) return d;
    }
  }
  if (flag == 0) return s;
  path.pop_back();
  return -1;
}
int main() {
  int n, m, k, u, v;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &u, &v);
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  vector<int> path;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (!visit[i]) {
      int d = dfs(i, path);
      if (d != -1) {
        for (int j = 0; j < g[d].size(); j++) parr[g[d][j]] = 1;
        int j;
        for (j = 0; j < path.size(); j++) {
          if (parr[path[j]] == 1) {
            break;
          }
        }
        for (; j < path.size(); j++) ans.push_back(path[j]);
        printf("%d\n", ans.size());
        for (j = 0; j < ans.size(); j++) printf("%d ", ans[j] + 1);
        break;
      }
    }
  }
  return 0;
}
