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 double eps = 1e-8;
const int mod = 1e9 + 7;
const int inf = 1061109567;
const int maxn = 1e5 + 5;
int head[maxn * 2], d[maxn], Dfs[maxn], ans[maxn], cnt = 0;
struct node {
  int to, nextt;
} e[maxn * 2];
int num = 0;
void add(int u, int v) {
  e[num].to = v;
  e[num].nextt = head[u];
  head[u] = num++;
}
int k, st, ed;
int dfs(int u, int de, int fa) {
  Dfs[u] = de;
  ans[de] = u;
  for (int i = head[u]; ~i; i = e[i].nextt) {
    int v = e[i].to;
    if (Dfs[v]) {
      if (de - Dfs[v] < k) {
        continue;
      }
      ed = de;
      st = Dfs[v];
      return 1;
    }
    if (dfs(v, de + 1, u)) return 1;
  }
  return 0;
}
int main() {
  int n, m;
  cin >> n >> m >> k;
  memset(head, -1, sizeof(head));
  int x, y;
  while (m--) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0, -1);
  cout << ed - st + 1 << endl;
  for (int i = st; i <= ed; i++) {
    cout << ans[i] << " ";
  }
}
