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: Wrong Answer)

#require 'pp'
require 'set'

class CycleError < StandardError; end

n = gets.chomp!.to_i
graph = {}
(n-1).times do |i|
    a, b = gets.split.map(&:to_i) 
    graph[a] ||= {}
    graph[a][b] = 1
    graph[b] ||= {}
    graph[b][a] = 1
end

#pp graph
dist3 = Set.new
(n-1).times do |i|
    x = [[i+1, 0]]
    path = Array.new(n, -1)
    path[i] = nil

    while x.empty?.!
        t, dist = x.pop
        path[t-1] = dist
        if dist == 3
            dist3 << [t, i+1].minmax
            next
        end
        graph[t].keys.each do |k, _|
            next if path[k-1] != -1
            x.push([k, dist + 1])
        end
    end

    #pp path
end

#pp dist3
graph2 = {}
dist3.each do |n1, n2|
    graph2[n1] ||= SortedSet.new
    graph2[n1] << n2
    graph2[n2] ||= SortedSet.new
    graph2[n2] << n1
end

#pp graph2

begin
    path = Array.new(n, -1)
    (n-1).times do |i|
        next unless graph2[i+1]
        next if path[i] != -1
        x = [
            [
                i+1,
                -1, # before
            ]
        ]
        path[i] = i

        while x.empty?.!
            t, before = x.pop
            #pp t
            path[t-1] = 1
            graph2[t].each do |k, _|
                next if before > 0 && k == before
                
                if path[k-1] > 0
                    # 循環
                    raise CycleError
                end

                x.push([k, t])
            end
        end

        #pp path
    end
rescue CycleError
    puts "-1"
    exit 0
end

# 循環しなかった
baisu = []
baisu_x = []
1.upto(n) do |i|
    if i % 3 == 0
        baisu << i
    else
        baisu_x << i
    end
end

#pp baisu
#pp baisu_x
#pp graph2
rem = []
ans = Array.new(n)
n.times do |i|
    unless graph2[i+1]
        rem << i
        next
    end

    #pp "==="
    #pp graph2
    #pp i+1
    #pp graph2[i+1]

    if graph2[i+1].length == 2
        ans[i] = baisu.shift
    else
        rem << i
    end
end

x = baisu + baisu_x
rem.each do |i|
    ans[i] = x.shift
end

puts ans.join(" ")