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)

if !ENV['RUBY_THREAD_VM_STACK_SIZE']
    require 'rubygems'
    RUBY=Gem.ruby
    exec({'RUBY_THREAD_VM_STACK_SIZE'=>'100000000'},RUBY,$0)
end

N = gets.to_i
WHITE = 0
BLACK = 1

G = Hash.new { |h, k| h[k] = [] }

(N - 1).times do
  a, b = gets.split.map(&:to_i)

  G[a] << b
  G[b] << a
end

COLOR = Hash.new

def dfs(v, color = WHITE, visited = Hash.new(false))
  visited[v] = true
  COLOR[v] = color

  G[v].each do |u|
    next if visited[u]

    if color == WHITE
      dfs(u, BLACK, visited)
    else
      dfs(u, WHITE, visited)
    end
  end
end

dfs(1)
nums = Hash.new { |h, k| h[k] = [] }

1.upto(N) do |n|
  nums[n % 3] << n
end

black_cnt = COLOR.values.count(BLACK)
white_cnt = COLOR.values.count(WHITE)
ans = []

if black_cnt <= N / 3
  1.upto(N) do |n|
    if COLOR[n] == BLACK
      ans[n] = nums[0].pop
    else
      if nums[1].empty?
        ans[n] << nums[2].pop
      else
        ans[n] << nums[1].pop
      end
    end
  end
elsif white_cnt <= N / 3
  1.upto(N) do |n|
    if COLOR[n] == WHITE
      ans[n] = nums[0].pop
    else
      if nums[1].empty?
        ans[n] << nums[2].pop
      else
        ans[n] << nums[1].pop
      end
    end
  end
else
  1.upto(N) do |n|
    if COLOR[n] == WHITE
      ans[n] = nums[1].pop if nums[1].size > 0
    else
      ans[n] = nums[2].pop if nums[2].size > 0
    end
  end

  1.upto(N) do |n|
    ans[n] = nums[0].pop if ans[n].nil?
  end
end

puts ans[1..-1].join(' ')