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;
vector<vector<int> > v;
bool visit[100005];
int val[100005];
vector<int> ans;
int r;
int rec(int cur, int len, int k, int last) {
  if (visit[cur]) {
    if (val[last] - val[cur] + 1 >= k) {
      r = val[last] - val[cur] + 1;
      return cur;
    } else {
      return 0;
    }
  }
  visit[cur] = 1;
  val[cur] = len;
  for (int i = 0; i < v[cur].size(); i++) {
    int rt = rec(v[cur][i], len + 1, k, cur);
    if (rt == -1) return -1;
    if (rt != 0)
      if (rt != -1) {
        ans.push_back(v[cur][i]);
        if (rt == cur) {
          return -1;
        } else
          return rt;
      }
  }
  visit[cur] = 0;
  return 0;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  v.resize(n + 2);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  memset(visit, 0, sizeof(visit));
  int rt = rec(1, 0, k + 1, -1);
  if (rt != 0) {
    cout << r << "\n";
    for (int i = ans.size() - 1; i >= 0; i--) cout << ans[i] << " ";
  }
  return 0;
}
