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 INF = (-1u) / 2;
const long long int INF2 = (-1ull) / 2;
int a, b, i, us[1011000], j, k, n, m, timer = 0, l, r, x, y;
int c[1011000], cnt = 0, fl = 0, a2, a3 = -1000000, ans = 0;
vector<int> g[101001], res;
int dfs(int v, int par) {
  if (par != -1)
    us[v] = us[par] + 1;
  else
    us[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    int k;
    if (to == par) continue;
    if (us[to] && us[v] - us[to] + 1 >= x + 1) {
      res.push_back(v);
      return to;
    } else if (!us[to]) {
      k = dfs(to, v);
      if (k != v && k != 0) {
        res.push_back(v);
        return k;
      } else if (k != 0) {
        res.push_back(v);
        cout << res.size() << endl;
        for (i = res.size() - 1; i >= 0; i--) cout << res[i] << " ";
        exit(0);
      }
    }
  }
  return 0;
}
int main() {
  scanf("%d%d%d", &n, &m, &x);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (i = 1; i <= n; i++) {
    if (!us[i]) k = dfs(i, -1);
  }
  return 0;
}
