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)

n = gets.to_i

count = Array.new(n + 1, 0)
nodes = Array.new(n + 1).map{ |x| Array.new }
count3 = Array.new(n + 1, 0)
#nodes3 = Array.new(n + 1).map{ |x| Array.new }

(n-1).times{
	a,b = gets.split.map &:to_i
	nodes[a] << b
	nodes[b] << a
	count[a] += 1
	count[b] += 1
}

# 端点
ind = count.index{ |x| x == 1 }

temp = [ind]
nodes_len = Array.new(n + 1).map{|x| Array.new }
routed = Array.new(n + 1, false)
routed[ind] = true

nodes_len[0] = [ind]
len = 1

while temp.size > 0
	next_temp = []
	
	temp.each{ |j|
		nodes[j].each{ |i|
			if ! routed[i] then
				next_temp << i
				routed[i] = true
			end
		}
	}
	
	if len >= 3 then
		add = nodes_len[len - 3].size
		next_temp.each{ |i|
			count3[i] += add
		}
		
		add = next_temp.size
		nodes_len[len - 3].each{ |i|
			count3[i] += add
		}
	end

	nodes_len[len] = next_temp
	
	temp = next_temp
	len += 1
end

count3_sort = count3[1..-1].map.with_index{ |x, i| [x,i] }.sort{|a,b| b[0] <=> a[0] }
count3x = n / 3

ppp1 = count3_sort[count3x][0]
ppp2 = count3_sort[count3x + count3x][0]

if (n % 3 == 0 || n % 3 == 1) && ppp1 > n - n / 3  then
	puts -1
elsif n % 3 == 1 && ppp2 > n - n / 3 - 1
	puts -1
elsif n % 3 == 2 && ppp1 > n - n / 3 - 2
	puts -1
else
	result = Array.new(n)

	pp = (1..n).to_a.sort{|x| x % 3}

	count3_sort.map{ |x| x[1] }.each_with_index{ |i,j|
		result[i] = pp[j]
	}
	
	puts result.join(" ")
end