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 
def cycle(G, K):
	path = [1]
	pos = [-1 for i in range(len(G) + 1)]
	while True:
		v = path[-1]
		pos[v] = len(path) - 1
		#print(path, v, G[v])
		for u in G[v]:
			if pos[u] == -1:
				path.append(u)
				break
			elif len(path) - pos[u] > K:
				return path[pos[u]::]

def main():
	N, M, K = [int(i) for i in input().split()]
	G = {i: set() for i in range(1, N+1)}
	for m in range(M):
		i, j = [int(i) for i in input().split()]
		G[i].add(j)
		G[j].add(i)
	path = cycle(G,K)
	print(len(path))
	temp = ' '.join(str(v) for v in path)
	#resp = temp[::len(temp)-1]
	print(temp)


main()

# 1506712255730
