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 MAX = 100010;
int n, m, k, x, y, d[MAX], p[MAX], pointer = 1, par, len;
bool v[MAX], cycle[MAX];
vector<int> a[MAX], answer;
void DFS(int start) {
  v[start] = true;
  int current;
  for (int i = 0; i < a[start].size(); i++) {
    current = a[start][i];
    if (!v[current]) {
      d[current] = (d[start] + 1);
      if (d[current] > d[pointer]) {
        pointer = current;
      }
      p[current] = start;
      DFS(current);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  DFS(1);
  answer.push_back(pointer);
  par = p[pointer];
  while (par != 1) {
    answer.push_back(par);
    par = p[par];
  }
  answer.push_back(1);
  for (int i = 0; i < a[pointer].size(); i++) {
    cycle[a[pointer][i]] = true;
  }
  for (int i = answer.size() - 1; i >= 0; i--) {
    if (cycle[answer[i]]) {
      len = i;
      break;
    }
  }
  cout << len + 1 << "\n";
  for (int i = 0; i <= len; i++) {
    cout << answer[i] << " ";
  }
  cout << "\n";
  return 0;
}
