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 DFS(G, k):
    d = [0] * (len(G) + 1)
    mark = [0] * (len(G) + 1)
    pi = [0] * (len(G) + 1)
    back_edge = [0] * 2
    back_edge[0] = back_edge[1] = -1

    for i in range(1 ,len(G)):
        if(mark[i] == 0):
            pi[i] = -1
            DFS_VISIT(G, i, mark, d, pi, k, back_edge)
        if(back_edge[0] != -1 and back_edge [1] != -1):
            break

    current = back_edge[0]
    cycle = []
    while(current != pi[back_edge[1]]):
        cycle.append(current)
        current = pi[current]

    return cycle

def DFS_VISIT(G, u, mark, d, pi, k, back_edge):
     current = u
     while(current != -1):
        mark[current] = 1

        done = 1
        for v in G[current]:
            if mark[v] == 0:
                done = 0
                pi[v] = current
                d[v] = d[current] + 1
                current = v
                break
            else:
                if d[current] - d[v] + 1 >= k + 1:
                    back_edge[0] = current
                    back_edge[1] = v

        if(done == 1): current = pi[current]

def main():
    n, m, k = [int(i) for i in input().split()]
    G = {i: set() for i in range(1, n + 1)}
    for j in range(m):
        u, v = [int(d) for d in input().split()]
        G[u].add(v)
        G[v].add(u)
    cycle = DFS(G, k)
    print(len(cycle))
    s = ' '.join(str(u) for u in cycle)
    print(s)

main()
