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> q[100002];
int ans[100002];
int flag, v[100002];
int k;
void dfs(int len, int x, int f) {
  int i;
  v[x] = len;
  ans[len] = x;
  for (i = 0; i < q[x].size(); i++) {
    int j = q[x][i];
    if (j == f) continue;
    if (v[j] != -1) {
      if (flag) break;
      if (len + 1 - v[j] >= k + 1) {
        printf("%d\n", len + 1 - v[j]);
        for (i = v[j]; i <= len; i++) {
          if (i != v[j]) printf(" ");
          printf("%d", ans[i]);
        }
        printf("\n");
        flag = 1;
        return;
      }
      continue;
    }
    dfs(len + 1, j, x);
  }
}
int main() {
  int n, m;
  while (scanf("%d%d%d", &n, &m, &k) != EOF) {
    while (m--) {
      int a, b;
      scanf("%d%d", &a, &b);
      q[a].push_back(b);
      q[b].push_back(a);
    }
    memset(v, -1, sizeof(v));
    dfs(0, 1, -1);
  }
  return 0;
}
