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;
bool v[100001];
int d[100001];
vector<int> ad[100001];
stack<int> ans;
int ve;
vector<int> te;
int dfs(int no, int t) {
  v[no] = 1;
  d[no] = t;
  ans.push(no);
  ve = no;
  for (int i = 0; i < ad[no].size(); i++) {
    if (v[ad[no][i]] == 0) {
      dfs(ad[no][i], t + 1);
      return 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k, t1, t2;
  cin >> n >> m >> k;
  memset(v, 0, sizeof(v));
  memset(d, 0, sizeof(0));
  for (int i = 0; i < m; i++) {
    cin >> t1 >> t2;
    ad[t1].push_back(t2);
    ad[t2].push_back(t1);
  }
  dfs(1, 1);
  int uy, ty = 100001;
  for (int i = 0; i < ad[ve].size(); i++) {
    if (d[ad[ve][i]] < ty) {
      ty = d[ad[ve][i]];
      uy = ad[ve][i];
    }
  }
  while (!ans.empty()) {
    te.push_back(ans.top());
    if (ans.top() == uy) {
      ans.pop();
      break;
    }
    ans.pop();
  }
  cout << te.size() << "\n";
  for (int i = 0; i < te.size(); i++) {
    cout << te[i] << " ";
  }
  cout << "\n";
  return 0;
}
