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 N = 100010;
int n, m, k, ord[N], cyc[N];
vector<int> v[N];
int find(int x) {
  for (int i = 0; i < int(v[x].size()); ++i) {
    int y = v[x][i];
    if (!ord[y]) return y;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 1, x, y; i <= m; ++i) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  ord[1] = 1;
  cyc[1] = 1;
  int sz = 1;
  int x = 1;
  int y = find(x);
  while (y) {
    ord[y] = ord[x] + 1;
    cyc[++sz] = y;
    x = y;
    y = find(x);
  }
  int mn = 1 << 30;
  for (int i = 0; i < int(v[x].size()); ++i)
    if (ord[v[x][i]] < mn) {
      mn = ord[v[x][i]];
      y = v[x][i];
    }
  cout << ord[x] - mn + 1 << '\n';
  for (int i = mn; i <= ord[x]; ++i) cout << cyc[i] << ' ';
}
