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 long long mod = 1e9 + 7, siz = 1e6 + 5;
vector<long long> g[siz];
long long t, n, m, k, dist[siz], a[siz];
bool vis[siz];
stack<long long> path;
void go(long long u) {
  path.push(u);
  vis[u] = 1;
  for (auto v : g[u]) {
    if (vis[v]) {
      if (abs(dist[v] - dist[u]) >= k) {
        vector<long long> out;
        while (path.top() != v) {
          out.push_back(path.top());
          path.pop();
        }
        out.push_back(path.top());
        cout << out.size() << endl;
        for (auto x : out) cout << x << " ";
        exit(0);
      }
    } else {
      dist[v] = 1 + dist[u];
      go(v);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  go(1);
}
