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 MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
long long int min(long long int a, long long int b) { return (a < b) ? a : b; }
long long int gcd(long long int a, long long int b) {
  if (b == 1)
    return 1;
  else if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int n, m, k;
vector<int> adj[100005];
bool vis[100005];
bool fl = false;
vector<int> v;
void dfs(int s) {
  if (fl) return;
  vis[s] = true;
  v.push_back(s);
  vector<int>::iterator i;
  for (i = adj[s].begin(); i < adj[s].end(); i++) {
    if (!vis[*i])
      dfs(*i);
    else {
      if (fl) return;
      int j;
      int ctr = 0;
      for (j = 0; j < v.size(); j++) {
        if (v[j] == *i) {
          ctr++;
          break;
        }
      }
      if ((ctr == 1) && (v.size() - j > k)) {
        fl = true;
        cout << v.size() - j << '\n';
        int k1;
        for (k1 = j; k1 < v.size(); k1++) {
          cout << v[k1] << " ";
        }
        cout << '\n';
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  int i, j, l;
  for (i = 0; i < m; i++) {
    cin >> j >> l;
    adj[j].push_back(l);
    adj[l].push_back(j);
  }
  memset(vis, 0, sizeof(vis));
  vector<int> v;
  for (i = 1; i < n + 1; i++) {
    if (fl) return 0;
    if (!vis[i]) {
      v.clear();
      dfs(i);
    }
  }
  return 0;
}
