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 int maxn = 2e5 + 1;
vector<int> g[maxn];
int n, m, k, num[maxn], pre[maxn], maxx = 0, flag = 0;
void dfs(int u, int fa, int deep) {
  num[u] = deep;
  pre[u] = fa;
  for (int to : g[u]) {
    if (flag) return;
    if (to == fa) continue;
    if (num[to] > 0) {
      if (num[u] - num[to] + 1 >= k + 1) {
        printf("%d\n", num[u] - num[to] + 1);
        while (u != to) {
          printf("%d ", u);
          u = pre[u];
        }
        printf("%d", to);
        flag = 1;
        return;
      }
      continue;
    }
    dfs(to, u, deep + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, u, v; i <= m; ++i) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 1);
}
