263_D. Cycle in Graph
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 
```python
n, m, k = map(int, input().split())
d = {}
for _ in range(m):
    u, v = map(int, input().split())
    if u not in d:
        d[u] = [v]
    else:
        d[u].append(v)
    if v not in d:
        d[v] = [u]
    else:
        d[v].append(u)


# print(d)
stack = []
not_visited = set(range(1, n+1))

while not_visited:
    if not stack:
        stack.append((not_visited.pop(), 1, 0))
        path = []
        visited = {}
        c = 0
    p, l, parent = stack.pop()

    path.append(p)
    visited[p] = c
    c += 1
    if p in not_visited:
        not_visited.remove(p)
    f = False
    for x in d[p]:
        if x not in visited:
            stack.append((x, l+1, p))
        else:

            if c - visited[x] > k:
                path = path[visited[x]:]
                f = True
                break
    if f:
        break
print(len(path))
print(*path)
```