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 long MOD = 1e9 + 7;
const int iinf = 1 << 29;
const long long llinf = 1ll << 60;
const double PI = 3.14159265;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long mrand(long long B) { return (unsigned long long)rng() % B; }
long long cross(pair<long long, long long> p1, pair<long long, long long> p2) {
  return p1.first * p2.second - p2.first * p1.second;
}
bool online(pair<long long, long long> a, pair<long long, long long> b,
            pair<long long, long long> c) {
  pair<long long, long long> p1{b.first - a.first, b.second - a.second};
  pair<long long, long long> p2{c.first - a.first, c.second - a.second};
  return cross(p1, p2) == 0;
}
void z_func(vector<int> &z, vector<int> &s, int n) {
  int L = 0, R = 0;
  for (int i = 1; i < n; i++) {
    if (i > R) {
      L = R = i;
      while (R < n && s[R - L] == s[R]) R++;
      z[i] = R - L;
      R--;
    } else {
      int k = i - L;
      if (z[k] < R - i + 1)
        z[i] = z[k];
      else {
        L = i;
        while (R < n && s[R - L] == s[R]) R++;
        z[i] = R - L;
        R--;
      }
    }
  }
}
const int MXN = 1e5 + 1;
vector<int> g[MXN];
int depth[MXN];
int n, m, k;
vector<int> ans;
void print_ans(int start) {
  int i = 0;
  for (; i < ((int)(ans).size()); ++i)
    if (ans[i] == start) break;
  cout << ((int)(ans).size()) - i << "\n";
  for (; i < ((int)(ans).size()); ++i) cout << ans[i] + 1 << " ";
  cout << endl;
}
void dfs(int node, int d) {
  depth[node] = d;
  ans.push_back(node);
  if (((int)(ans).size()) >= k) {
    for (auto &e : g[node]) {
      if (depth[e] != -1 && d - depth[e] + 1 > k) {
        print_ans(e);
        exit(0);
      }
    }
  }
  for (auto &e : g[node]) {
    if (depth[e] != -1) continue;
    dfs(e, d + 1);
  }
  depth[node] = -1;
  ans.pop_back();
}
void work() {
  cin >> n >> m >> k;
  memset(depth, -1, sizeof(depth));
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v), g[v].push_back(u);
  }
  dfs(0, 1);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  work();
  return 0;
}
