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 Julia (0.5.0) (judge verdict: Accepted)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
	n = readline() |> parseInt
	e = [Int[] for i in 1:n]
	for i in 1:n-1
		a,b = readline() |> split |> parseMap
		push!(e[a],b)
		push!(e[b],a)
	end
	chk = -ones(Int,n)
	chk[1] = 0
	stack = Int[1]
	while !isempty(stack)
		now = pop!(stack)
		for i in e[now]
			if chk[i] == -1
				push!(stack,i)
				chk[i]=(chk[now]+1)%2
			end
		end
	end
	ans = zeros(Int,n)
	z = sum(chk)
	sx = 1
	sy = 2
	sz = 3
	if sum(chk)<=div(n,3)
		for i in 1:n
			if chk[i]==1
				ans[i] = sz
				sz += 3
			else
				if sy<=n
					ans[i] = sy
					sy += 3
				elseif sx<=n
					ans[i] = sx
					sx += 3
				else
					ans[i] = sz
					sz += 3
				end
			end
		end
	elseif sum(chk)>2*div(n,3)
		for i in 1:n
			if chk[i]==0
				ans[i] = sz
				sz += 3
			else
				if sy<=n
					ans[i] = sy
					sy += 3
				elseif sx<=n
					ans[i] = sx
					sx += 3
				else
					ans[i] = sz
					sz += 3
				end
			end
		end
	else
		for i in 1:n
			if chk[i]==0
				if sx<=n
					ans[i] = sx
					sx += 3
				else
					ans[i] = sz
					sz += 3
				end
			else
				if sy<=n
					ans[i] = sy
					sy += 3
				else
					ans[i] = sz
					sz += 3
				end
			end
		end
	end
	if n == 2
		println("1 2")
	else
		for i in 1:n-1
			print(ans[i], " ")
		end
		println(ans[n])
	end
end

main()