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>
const long long inf = 1e9 + 7;
const long double eps = 1e-9;
const long long N = 1e6 + 5;
const long long M = 1e3 + 5;
using namespace std;
void ret(string s) {
  cout << s;
  exit(0);
}
int n, m, k, v1, v2, len, h;
int num[N], ans[N];
vector<int> es[N];
bool u[N];
void dfs(int v) {
  int mn = N;
  bool wa = false;
  num[v] = ++len;
  ans[len] = v;
  u[v] = true;
  for (int i = 0; i < es[v].size(); i++) {
    h = es[v][i];
    if (!u[h]) {
      dfs(h);
      wa = true;
    } else {
      mn = min(mn, num[h]);
    }
  }
  if (!wa) {
    cout << len - mn + 1 << "\n";
    for (int i = mn; i <= len; i++) cout << ans[i] << " ";
    exit(0);
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> v1 >> v2;
    es[v1].push_back(v2);
    es[v2].push_back(v1);
  }
  dfs(1);
  return 0;
}
