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;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
bool visited[100005];
int par[100005];
int n, m, k, fin = -1;
vector<int> v[100005];
int len1 = 0;
void dfs(int x, int prev, int len, int st) {
  if (fin != -1) return;
  visited[x] = true;
  par[x] = prev;
  if (len >= k) {
    for (int i = 0; i < v[x].size(); i++) {
      int curr = v[x][i];
      if (curr == st) fin = x;
    }
  }
  if (fin != -1) return;
  for (int i = 0; i < v[x].size(); i++) {
    int curr = v[x][i];
    if (!visited[curr]) {
      len1++;
      dfs(curr, x, len + 1, st);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    cnt++;
    fill(visited, visited + n, false);
    fill(par, par + n, 0);
    len1 = 0;
    dfs(i, 0, 0, i);
    if (fin != -1) {
      break;
    }
  }
  vector<int> res;
  for (int i = fin; i != 0; i = par[i]) {
    res.push_back(i + 1);
  }
  if (cnt < 2) res.push_back(1);
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] << " ";
  }
  cout << endl;
  return 0;
}
