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;
constexpr long long maxn = 2e5 + 5, inf = 0x3f3f3f3f;
vector<long long> e[maxn];
long long n, m, k;
long long vis[maxn], pos;
void dfs(long long u, long long tim) {
  vis[u] = tim;
  for (auto v : e[u]) {
    if (!vis[v]) {
      dfs(v, tim + 1);
      break;
    }
    if (vis[u] >= vis[v] + k) {
      cout << vis[u] - vis[v] + 1 << endl;
      cout << u << " ";
      pos = v;
      return;
    }
  }
  if (pos != -1) cout << u << " ";
  if (pos == u) pos = -1;
}
signed main() {
  cin >> n >> m >> k;
  for (long long i = 1; i <= m; i++) {
    long long u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 1);
}
