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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local n = io.read("*n")
local edge = {}
local num = {}
local len = {}
local lim, use = {}, {}
lim[1] = math.ceil(n / 3)
lim[2] = math.ceil((n - 1) / 3)
use[1], use[2], use[3] = 0, 0, 0

for i = 1, n do
  edge[i] = {}
  num[i] = 0
  len[i] = -1
end
for i = 1, n - 1 do
  local a, b = io.read("*n", "*n")
  edge[a][b], edge[b][a] = true, true
end
local tasks = {1}
len[1] = 0
num[1] = 1
use[1] = 1
for i = 1, n do
  local src = tasks[i]
  local srcnum = num[src] % 3
  for dst, _u in pairs(edge[src]) do
    if len[dst] == -1 then
      len[dst] = len[src] + 1
      if len[dst] % 2 == 0 then
        if use[2] < lim[2] then
          use[2] = use[2] + 1
          num[dst] = use[2] * 3 - 1
        else
          use[3] = use[3] + 1
          num[dst] = use[3] * 3
        end
      else
        if use[1] < lim[1] then
          use[1] = use[1] + 1
          num[dst] = use[1] * 3 - 2
        else
          use[3] = use[3] + 1
          num[dst] = use[3] * 3
        end
      end
      table.insert(tasks, dst)
    end
  end
end
print(table.concat(num, " "))