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;
vector<int> v[100100];
bool fl = 0;
int n, m, st[100100], k;
int was[100100], in_dfs[100100];
bool cmp(int A, int B) {
  if (st[A] > st[B]) return 1;
  return 0;
}
int dfs(int x, int t) {
  if (fl == 1) return -1;
  was[x] = 1;
  in_dfs[x] = t;
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (!was[to]) {
      int e = dfs(to, t + 1);
      if (e == -1 && fl == 0) continue;
      if (e == -1 && fl == 1) return -1;
      if (e == x) {
        fl = 1;
        return -1;
      }
      printf("%d ", x);
      return e;
    }
    if (in_dfs[to] != -1) {
      if (t - in_dfs[to] + 1 >= k + 1) {
        printf("%d\n", t - in_dfs[to] + 1);
        printf("%d %d ", to, x);
        return to;
      }
    }
  }
  in_dfs[x] = -1;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int aa, bb;
    scanf("%d%d", &aa, &bb);
    v[aa].push_back(bb);
    v[bb].push_back(aa);
    st[aa]++;
    st[bb]++;
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end(), cmp);
  for (int i = 1; i <= n && fl == 0; i++)
    if (was[i] == 0) dfs(i, 0);
  if (fl == 0)
    while (1 == 1) n = n;
  return 0;
}
