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-9;
const int N = 100024;
vector<int> edge[N];
int n, m, k;
bool vis[N];
int beg_pos;
bool dfs(int pos, int cnt) {
  int size = edge[pos].size();
  if (cnt >= k) {
    for (int i = 0; i < size; ++i) {
      if (edge[pos][i] == beg_pos) {
        printf("%d\n", cnt);
        return 1;
      }
    }
  }
  for (int i = 0; i < size; ++i) {
    int tmp = edge[pos][i];
    if (vis[tmp] == 1) continue;
    vis[tmp] = 1;
    if (dfs(tmp, cnt + 1) == 1) {
      printf("%d ", tmp);
      return 1;
    }
    vis[tmp] = 0;
  }
  return 0;
}
int main() {
  while (scanf("%d%d%d", &n, &m, &k) != EOF) {
    ++k;
    (memset(vis, 0, sizeof(vis)));
    for (int i = 0; i < n + 1; ++i) edge[i].clear();
    for (int i = 0; i < m; ++i) {
      int a, b;
      scanf("%d%d", &a, &b);
      edge[a].push_back(b);
      edge[b].push_back(a);
    }
    bool flag = 0;
    for (int i = 1; i < n + 1; ++i) {
      beg_pos = i;
      vis[i] = 1;
      if (dfs(i, 1)) {
        printf("%d\n", i);
        flag = 1;
        break;
      }
      vis[i] = 0;
    }
    if (!flag) puts("-1");
  }
  return 0;
}
