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 
from sys import stdin
from collections import *


class graph:
    # initialize graph
    def __init__(self, gdict=None):
        self.gdict = [[] for _ in range(n + 1)]

    # add edge
    def addEdge(self, node1, node2, w=None):
        self.gdict[node1].append(node2)
        self.gdict[node2].append(node1)

    def dfsUtil(self, v):
        stack, self.visit = v, [-1] * (n + 1)
        ans, d = [], 0

        while (stack):
            ans.append(stack)
            self.visit[stack] = d
            d += 1

            for i in self.gdict[stack]:
                if self.visit[i] == -1:
                    stack = i
                    break
            else:
                be = ans.index(min(self.gdict[stack], key=lambda x: self.visit[x]))
                print('%d\n%s' % (len(ans) - be, ' '.join(map(str, ans[be:]))))
                exit()


rints = lambda: [int(x) for x in stdin.readline().split()]
n, m, k = rints()
g = graph()
for _ in range(m):
    u, v = rints()
    g.addEdge(u, v)

g.dfsUtil(1)
