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> v[100010];
vector<int> ans;
int k;
int n;
int m;
int st;
int vis[100010];
bool dfs(int i, int dep) {
  ans.push_back(i);
  vis[i] = dep;
  for (int j = 0; j < v[i].size(); ++j) {
    if (vis[v[i][j]] == -1) {
      if (dfs(v[i][j], dep + 1)) return true;
    } else if (vis[v[i][j]] + k <= dep) {
      st = v[i][j];
      return true;
    }
  }
  return false;
}
int main() {
  while (scanf("%d%d%d", &n, &m, &k) == 3) {
    ans.clear();
    memset(vis, 0xff, sizeof(vis));
    for (int i = 0; i <= n; ++i) v[i].clear();
    for (int i = 0; i < m; ++i) {
      int a, b;
      scanf("%d%d", &a, &b);
      v[a].push_back(b);
      v[b].push_back(a);
    }
    st = -1;
    if (false == dfs(1, 0)) puts("error");
    int i = 0;
    while (i < ans.size()) {
      if (ans[i] == st) break;
      i++;
    }
    printf("%d\n", ans.size() - i);
    for (; i < ans.size(); ++i) {
      printf("%d ", ans[i]);
    }
    puts("");
  }
  return 0;
}
