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;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline string get() {
  char ch[1000010];
  scanf("%s", ch);
  return ch;
}
const int MAX_LG = 40;
const long long maxn = 1e5 + 100;
const long long base = 29;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 100;
vector<long long> g[maxn];
bool mark[maxn];
bool N[maxn];
vector<long long> path;
inline void dfs(long long v) {
  mark[v] = true;
  path.push_back(v);
  for (auto u : g[v]) {
    if (mark[u]) continue;
    dfs(u);
    break;
  }
}
int32_t main() {
  long long n = in(), m = in(), k = in();
  for (long long i = 0; i < m; i++) {
    long long v = in() - 1, u = in() - 1;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0);
  long long en = path.back();
  for (auto u : g[en]) N[u] = true;
  long long cur = 0;
  for (auto v : path) {
    if (N[v]) {
      cout << (long long)path.size() - cur << "\n";
      break;
    }
    cur++;
  }
  for (long long j = cur; j < path.size(); j++) cout << path[j] + 1 << " ";
  cout << "\n";
}
