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 MAXN = 1e+5 + 50;
int n, m, k;
vector<int> g[MAXN];
vector<int> ans;
int vis[MAXN];
int deep[MAXN];
int start, found;
void init() {
  memset(vis, 0, sizeof(vis));
  memset(deep, 0, sizeof(deep));
  for (int i = 0; i <= MAXN - 1; i++) {
    g[i].clear();
  }
  ans.clear();
  start = found = 0;
}
void dfs(int pos, int father, int len) {
  vis[pos] = 1;
  deep[pos] = len;
  for (int i = 0; i < g[pos].size(); i++) {
    int temp = g[pos][i];
    if (temp == father) {
      continue;
    }
    if (vis[temp] && deep[pos] - deep[temp] + 1 >= k + 1) {
      start = temp;
      ans.push_back(pos);
      return;
    }
    if (!vis[temp]) {
      dfs(temp, pos, len + 1);
    }
    if (start) {
      if (found) {
        return;
      }
      if (pos == start) {
        found = 1;
      }
      ans.push_back(pos);
      return;
    }
  }
}
int main() {
  while (cin >> n >> m >> k) {
    init();
    int from, to;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &from, &to);
      g[from].push_back(to);
      g[to].push_back(from);
    }
    dfs(1, -1, 1);
    cout << ans.size() << endl;
    cout << ans[0];
    for (int i = 1; i < ans.size(); i++) {
      cout << " " << ans[i];
    }
    cout << endl;
  }
  return 0;
}
