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 INP, AM, REACHEOF;
char BUF[(1 << 12) + 1], *inp = BUF;
const long double PI = acos((long double)-1.0);
int n, m, k;
vector<int> ke[100111];
int res[100111], nRes;
int pos[100111];
int main() {
  while (scanf("%d%d%d", &n, &m, &k) == 3) {
    for (int i = (1), _b = (n); i <= _b; i++) ke[i].clear();
    memset(pos, 0, sizeof pos);
    for (int i = (1), _b = (m); i <= _b; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      ke[u].push_back(v);
      ke[v].push_back(u);
    }
    res[1] = 1;
    nRes = 1;
    pos[1] = 1;
    while (true) {
      int u = res[nRes];
      bool found = false;
      for (int i = 0, _a = (ke[u].size()); i < _a; i++) {
        int v = ke[u][i];
        if (!pos[v]) {
          pos[v] = nRes + 1;
          res[++nRes] = v;
          found = true;
          break;
        }
      }
      if (!found) break;
    }
    int first = nRes, u = res[nRes];
    for (int i = 0, _a = (ke[u].size()); i < _a; i++) {
      int v = ke[u][i];
      if (first = min(first, pos[v]))
        ;
    }
    printf("%d\n", nRes - first + 1);
    for (int i = (first), _b = (nRes); i <= _b; i++) printf("%d ", res[i]);
    puts("");
  }
}
