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<int> a[111111];
vector<int> path;
bool v[111111] = {0};
bool done[11111] = {0};
int dis[111111] = {0};
int n, m, k;
bool ok = false;
void dfs(int i, int deep) {
  if (v[i]) return;
  v[i] = true;
  path.push_back(i);
  dis[i] = deep;
  for (int j = 0; j < a[i].size() && !ok; j++) {
    if (v[a[i][j]] && !done[a[i][j]] && deep - dis[a[i][j]] >= k) {
      int pos = 0;
      while (path[pos] != a[i][j]) pos++;
      cout << deep - pos + 1 << endl;
      for (int k = pos; k <= deep; k++) {
        cout << path[k] << " ";
      }
      cout << endl;
      ok = true;
    } else {
      dfs(a[i][j], deep + 1);
    }
  }
  done[i] = true;
  path.pop_back();
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 0);
  return 0;
}
