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 long long int N = 1e5 + 1;
long long int mod = 998244353;
void normalize(long long int &a) { a = (a + mod) % mod; }
long long int inf = 10000000000000000;
vector<long long int> G[N];
vector<long long int> parent(N);
vector<long long int> visited(N);
vector<long long int> size(N, 0);
long long int modmul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a * b) % mod;
}
long long int fact(long long int n) {
  if (n == 0 || n == 1)
    return 1;
  else
    return modmul(n, fact(n - 1));
}
long long int s, e;
long long int q = 0;
long long int k;
void dfs(long long int v, long long int u) {
  visited[v] = 1;
  size[v] = size[u] + 1;
  parent[v] = u;
  for (auto y : G[v]) {
    if (y != u && visited[y] == 1 && size[v] - size[y] >= k) {
      s = y;
      e = v;
    } else if (visited[y] == 0)
      dfs(y, v);
  }
}
void solve() {
  long long int n, m;
  cin >> n >> m >> k;
  long long int k1, k2;
  for (long long int i = 0; i < m; i++) {
    cin >> k1 >> k2;
    G[k1].push_back(k2);
    G[k2].push_back(k1);
  }
  dfs(1, 1);
  long long int k3 = e;
  long long int count = 1;
  while (k3 != s) {
    k3 = parent[k3];
    count++;
  }
  cout << count << endl;
  cout << e << " ";
  while (e != s) {
    e = parent[e];
    cout << e << " ";
  }
}
signed main() {
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
}
