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 int N = 100100;
int n, m, k, p[N], niv[N], sol, tata;
vector<int> v[N];
void dfs(int nod) {
  if (sol) return;
  for (vector<int>::iterator it = v[nod].begin(); it != v[nod].end(); ++it)
    if (p[nod] != *it) {
      if (!p[*it]) {
        p[*it] = nod;
        niv[*it] = niv[nod] + 1;
        dfs(*it);
      } else {
        if (niv[nod] - niv[*it] >= k) {
          sol = nod;
          tata = *it;
          return;
        }
      }
    }
}
int main() {
  int i, a, b;
  cin >> n >> m >> k;
  for (i = 1; i <= m; ++i) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  p[1] = -1;
  dfs(1);
  cout << niv[sol] - niv[tata] + 1 << "\n";
  while (sol != tata) {
    cout << sol << " ";
    sol = p[sol];
  }
  cout << sol << "\n";
  return 0;
}
