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 
class Graph:
    def __init__(self, size=0):
        self.g = [[] for i in range(size)]
        self.size = size
        self.path = []

    def addEdge(self, u, v):
        try:
            self.g[u].append(v)
            self.g[v].append(u)
        except(IndexError):
            pass

    def DFSAux(self, v, k, start, root):
        d = 0
        visit = [root]

        while len(visit) > 0:
            r = visit.pop()
            if v[r] == -1:
                v[r] = d
                d += 1

                self.path.append(r)

                for e in self.g[r]:
                    if v[e] != -1:
                        if v[r] - v[e] + 1 >= k:
                            self.path = self.path[v[e]:v[r]+1]
                            return True
                    else:
                        visit.append(e)
        return False

    def findCycle(self,k):
        v = [-1 for i in range(self.size)]

        self.DFSAux(v,k,0,0)

if __name__ == '__main__':

    values = list(map(int,input().split()))

    g = Graph(values[0])
    m = values[1]

    while m > 0:
        m -= 1
        edges = list(map(int,input().split()))
        g.addEdge(edges[0]-1,edges[1]-1)

    g.findCycle(values[2]+1)

    print(len(g.path))
    for i in g.path:
        print(i+1,end=' ')
    print()

# 1506714347231
