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;
long long int n, m, k;
vector<long long int> v[100005], h(100005);
stack<long long int> second;
vector<long long int> ans;
void dfs(long long int x, long long int p) {
  h[x] = h[p] + 1;
  second.push(x);
  for (auto &y : v[x]) {
    if (ans.size() > 0) return;
    if (h[y] && h[x] - h[y] + 1 >= k + 1) {
      while (true) {
        ans.push_back(second.top());
        if (second.top() == y) break;
        second.pop();
      }
    } else if (!h[y])
      dfs(y, x);
  }
  second.pop();
}
void sol() {
  cin >> n >> m >> k;
  for (long long int i = 0; i < m; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans.size() << "\n";
  reverse(ans.begin(), ans.end());
  for (auto &x : ans) cout << x << " ";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tests = 1;
  while (tests--) {
    sol();
  }
}
