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>
#pragma comment(linker, "/STACK:26000000000")
using namespace std;
const int MAXN = 100005;
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
inline int triangle_area(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1);
}
int n, m, k, a, b;
vector<int> g[MAXN];
vector<char> cl;
vector<int> p;
int cycle_st, cycle_end;
void dfs(int v) {
  cl[v] = 1;
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (cl[to] == 0) {
      p[to] = v;
      dfs(to);
    } else if (cl[to] == 1 && to == 1) {
      cycle_end = v;
      cycle_st = to;
      vector<int> cycle;
      for (int h = cycle_end; h != cycle_st; h = p[h]) cycle.push_back(h);
      cycle.push_back(cycle_st);
      if ((int)cycle.size() >= k + 1) {
        printf("%d\n", (int)cycle.size());
        for (int h = (int)cycle.size() - 1; h >= 0; --h)
          printf("%d ", cycle[h]);
        exit(0);
      }
    }
  }
  cl[v] = 2;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  cl.resize(n + 1, 0);
  p.resize(n + 1, -1);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  return 0;
}
