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;
map<pair<int, int>, bool> d;
vector<int> a[100001];
int n, m, i, k, u, v, cnt, dd[100001], ds[100001], low[100001], st[100001], top,
    tr[100001], h[100001];
void DFS(int u, int p) {
  ds[u] = low[u] = ++cnt;
  st[++top] = u;
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (v == p || dd[v] == 1) continue;
    if (ds[v] == 0) {
      tr[v] = u;
      h[v] = h[u] + 1;
      DFS(v, u);
      low[u] = min(low[u], low[v]);
    } else
      low[u] = min(low[u], ds[v]);
  }
  if (low[u] == ds[u]) {
    int v;
    do {
      v = st[top--];
      dd[v] = 1;
      if (h[v] - h[u] >= k && d[make_pair(v, u)] == 1) {
        cout << h[v] - h[u] + 1 << '\n';
        do {
          cout << v << " ";
          v = tr[v];
        } while (v != tr[u]);
        exit(0);
      }
    } while (v != u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (i = 1; i <= m; i++) {
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
    d[make_pair(u, v)] = d[make_pair(v, u)] = 1;
  }
  for (i = 1; i <= n; i++)
    if (ds[i] == 0) DFS(i, 0);
}
