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;
int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[] = {0, 0, 1, -1, -1, 1, 1, -1};
int biton(int n, int pos) { return n = n | (1 << pos); }
int bitoff(int n, int pos) { return n = n & ~(1 << pos); }
bool bitcheck(int n, int pos) { return (bool)(n & (1 << pos)); }
inline int nxt() {
  int a;
  scanf("%d", &a);
  return a;
}
inline long long int NXT() {
  long long int a;
  scanf("%lld", &a);
  return a;
}
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
vector<int> graph[100010];
vector<int> ans;
int visited[100010];
int k;
int len;
bool dfs(int n, int depth) {
  if (visited[n] != -1) {
    if (depth - visited[n] >= k + 1) {
      len = depth - visited[n];
      return true;
    }
    return false;
  }
  visited[n] = depth;
  for (int i = 0; i < graph[n].size(); i++) {
    if (dfs(graph[n][i], depth + 1)) {
      ans.push_back(n);
      return true;
    }
  }
  return false;
}
int main() {
  int n = nxt(), m = nxt();
  k = nxt();
  for (int i = 0; i < m; i++) {
    int x = nxt();
    int y = nxt();
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  memset(visited, -1, sizeof(visited));
  dfs(1, 0);
  printf("%d\n", len);
  for (int i = 0; i < len; i++) printf("%d ", ans[i]);
}
