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 maxn = 100000 * 4;
struct node {
  int v, nxt;
} edge[maxn * 4];
int cnt = 0;
int head[maxn];
void add_Edge(int u, int v) {
  edge[cnt].v = v;
  edge[cnt].nxt = head[u];
  head[u] = cnt++;
}
int step = 0;
int k, flag = 0;
int sx, sy;
int answer[maxn], dfn[maxn];
int dfs(int u) {
  if (flag) return 1;
  dfn[u] = ++step;
  answer[step] = u;
  for (int i = head[u]; i != -1; i = edge[i].nxt) {
    int v = edge[i].v;
    if (dfn[v]) {
      if (abs(dfn[v] - dfn[u]) >= k) {
        flag = 1;
        sx = u;
        sy = v;
        return 1;
      }
    } else {
      dfs(v);
    }
  }
}
int main() {
  int n, m;
  scanf("%d%d%d", &n, &m, &k);
  cnt = 0;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= m; i++) {
    int xx, yy;
    scanf("%d%d", &xx, &yy);
    add_Edge(xx, yy);
    add_Edge(yy, xx);
  }
  dfs(1);
  printf("%d\n", dfn[sy] - dfn[sx] + 1);
  for (int i = dfn[sx]; i <= dfn[sy]; i++) printf("%d ", answer[i]);
}
