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 n, m, k;
vector<int> edge[100001];
int depth[100001];
stack<int> road;
void dfs(int v, int d) {
  if (!depth[v]) {
    road.push(v);
    depth[v] = d;
    for (vector<int>::iterator it = edge[v].begin(); it != edge[v].end(); it++)
      dfs(*it, d + 1);
    road.pop();
    depth[v] = 0;
  } else if (d - depth[v] > k) {
    printf("%d\n", d - depth[v]);
    while (!road.empty()) {
      printf("%d ", road.top());
      if (road.top() == v) break;
      road.pop();
    }
    printf("\n");
    exit(0);
  }
}
int main(void) {
  scanf("%d %d %d", &n, &m, &k);
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  memset(depth, 0, sizeof(int) * (n + 1));
  dfs(1, 1);
}
