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 long double eps = 1e-7;
const string alpha = "abcdefghijklmnopqrstuvwxyz";
int dx[8] = {+1, +1, +1, 0, 0, -1, -1, -1};
int dy[8] = {+1, 0, -1, +1, -1, +1, 0, -1};
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long sqr(long long x) { return x * x; }
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <class T>
inline void chmax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void chmin(T &x, T y) {
  x = min((x), (y));
}
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T powm(T a, T b, long long m) {
  T cnt = 1;
  while (b > 0) {
    if (b % 2 == 1) cnt = (cnt * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return cnt % m;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
const int MOD = 1000000007;
const int N = 100005;
void prec() {}
int n, m, k, col[N], par[N];
vector<int> ad[N], v;
void makeG() {
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    ad[u].push_back(v);
    ad[v].push_back(u);
  }
}
void dfs(int x) {
  col[x] = 1;
  for (auto &it : ad[x]) {
    if (!col[it])
      par[it] = x, dfs(it);
    else if (col[it] == 1 and it != par[x] and k) {
      int en = x;
      int cnt = 1;
      while (en != it) {
        cnt++;
        en = par[en];
      }
      if (cnt > k) {
        int st = x;
        v.push_back(st);
        while (st != it) {
          st = par[st];
          v.push_back(st);
        }
        k = 0;
        break;
      }
    }
  }
  col[x] = 2;
}
void run() {
  cin >> n >> m >> k;
  makeG();
  for (int i = 1; i <= n; ++i) {
    if (col[i] == 0 and k)
      dfs(i);
    else
      break;
  }
  cout << (long long)v.size() << "\n";
  for (auto &it : v) cout << it << " ";
}
int32_t main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int tc = 1;
  prec();
  for (long long i = 0; i < tc; ++i) run();
  return 0;
}
