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 MAX = 100010;
int P[MAX];
vector<int> E[MAX];
map<pair<int, int>, int> M;
int L = -1;
void dfs(int x) {
  for (int y : E[x]) {
    if (P[y] == 0) {
      P[y] = x;
      dfs(y);
      return;
    }
  }
  L = x;
}
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  while (m--) {
    int a, b;
    scanf("%d %d", &a, &b);
    E[a].push_back(b);
    E[b].push_back(a);
    M[pair<int, int>(a, b)] = 1;
    M[pair<int, int>(b, a)] = 1;
  }
  P[1] = -1;
  dfs(1);
  int length = 1, v = L;
  while (true) {
    v = P[v];
    length++;
    if (length > k && M[pair<int, int>(v, L)]) {
      printf("%d\n", length);
      for (int x = L; x != v; x = P[x]) {
        printf("%d ", x);
      }
      printf("%d\n", v);
      return (0);
    }
  }
}
