Problem:
Score : 500 points

Problem Statement

You are given a connected undirected simple graph, which has N vertices and M edges.
The vertices are numbered 1 through N, and the edges are numbered 1 through M.
Edge i connects vertices A_i and B_i.
Your task is to find a path that satisfies the following conditions:

The path traverses two or more vertices.

The path does not traverse the same vertex more than once.

A vertex directly connected to at least one of the endpoints of the path, is always contained in the path.

It can be proved that such a path always exists.
Also, if there are more than one solution, any of them will be accepted.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i < B_i \leq N

The given graph is connected and simple (that is, for every pair of vertices, there is at most one edge that directly connects them).

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Find one path that satisfies the conditions, and print it in the following format.
In the first line, print the count of the vertices contained in the path.
In the second line, print a space-separated list of the indices of the vertices, in order of appearance in the path.

Sample Input 1

5 6
1 3
1 4
2 3
1 5
3 5
2 4

Sample Output 1

4
2 3 1 4

There are two vertices directly connected to vertex 2: vertices 3 and 4.
There are also two vertices directly connected to vertex 4: vertices 1 and 2.
Hence, the path 2 → 3 → 1 → 4 satisfies the conditions.

Sample Input 2

7 8
1 2
2 3
3 4
4 5
5 6
6 7
3 5
2 6

Sample Output 2

7
1 2 3 4 5 6 7

Solution in LuaJIT (2.0.4) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local edge = {}
local asked = {}
for i = 1, n do
  edge[i] = {}
  asked[i] = false
end
for i = 1, m do
  local a, b = io.read("*n", "*n")
  edge[a][b], edge[b][a] = true, true
end

local pos1 = 1
local pos2 = next(edge[1])

local forward = {pos1}
local backward = {pos2}
asked[pos1], asked[pos2] = true, true
local hasdst = true
while hasdst do
  asked[pos1] = true
  hasdst = false
  for dst, _u in pairs(edge[pos1]) do
    if not asked[dst] then
      pos1 = dst
      table.insert(forward, dst)
      hasdst = true
      break
    end
  end
end
hasdst = true
while hasdst do
  asked[pos2] = true
  hasdst = false
  for dst, _u in pairs(edge[pos2]) do
    if not asked[dst] then
      pos2 = dst
      table.insert(backward, dst)
      hasdst = true
      break
    end
  end
end
for i = 1, math.floor(#backward / 2) do
  backward[i], backward[#backward + 1 - i] = backward[#backward + 1 - i], backward[i]
end
for i = 1, #forward do
  table.insert(backward, forward[i])
end
print(#backward)
print(table.concat(backward, " "))