Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i.

Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition:

For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3.

Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j.

Help Takahashi by finding a permutation that satisfies the condition.

Constraints

2\leq N\leq 2\times 10^5

1\leq a_i, b_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}

Output

If no permutation satisfies the condition, print -1.

Otherwise, print a permutation satisfying the condition, with space in between.
If there are multiple solutions, you can print any of them.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

N = gets.to_i
@edges = Array.new(N+1){ Array.new }
@edges[0].push(1)
(N-1).times do 
  a, b = gets.split.map(&:to_i)
  @edges[a].push(b)
  @edges[b].push(a)
end

@pairs = []

@qq = [
  [1, nil, nil, nil]
]

def trav node, parent, gparent, ggparent
  if ggparent
    @pairs.push([node, ggparent])
  end
  if gparent
    @edges[gparent].each do |nn|
      next if nn == parent
      @pairs.push([node, nn])
    end
  end
  @edges[node].each do |child|
    next if child == parent
    @qq.push([child, node, parent, gparent])
  end
end

while q = @qq.shift do
  trav(*q)
end
result = Array.new N
counts = {}
@pairs.each do |pair|
  counts[pair[0]] ||= 0
  counts[pair[1]] ||= 0
  counts[pair[0]] += 1
  counts[pair[1]] += 1
end
used = 0
counts.each do |k,v|
  if v > 1
    used += 3
    if used > N
      break
    end
    result[k-1] = used
  end
end
@pairs.each do |pair|
  next if result[pair[0]-1] || result[pair[1]-1]
  used += 3
  result[pair[0]-1] = used
end
raise if used > N
nused = 0
p result
result = result.map do |x|
  if x
    x
  else
    if nused == 0 || nused % 3 != 2
      nused += 1
    else
      nused += 2
    end
    nused
  end
end

puts result.join(' ')