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 maxn = 1e5 + 20;
long long n, m, k, tmp1, tmp2, H[maxn], flag, last, r;
vector<long long> adj[maxn], ans;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (long long i = 0; i < m; i++) {
    cin >> tmp1 >> tmp2;
    adj[tmp1].push_back(tmp2);
    adj[tmp2].push_back(tmp1);
  }
  ans.push_back(1);
  H[1] = 1;
  last = 1;
  while (!flag) {
    flag = 1;
    for (long long i = 0; i < adj[last].size(); i++) {
      long long u = adj[last][i];
      if (!H[u]) {
        H[u] = H[last] + 1;
        last = u;
        ans.push_back(u);
        flag = 0;
        break;
      }
    }
  }
  r = maxn;
  for (long long i = 0; i < adj[last].size(); i++) {
    r = min(r, H[adj[last][i]]);
  }
  r--;
  cout << ans.size() - r << endl;
  for (long long i = r; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
