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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

from collections import deque
n = int(input())
e = [[] for _ in range(n)]
AB = [list(map(int, input().split())) for _ in range(n - 1)]
AB = [(a -1, b - 1) for a, b in AB]

for itr, (a, b) in enumerate(AB):
  e[a].append(b)
  e[b].append(b)
  
colors = [-1] * n

# ２色で塗分けする。
q = deque()
# 現在見ている頂点、親の色
q.append((0, 0))
while q:
  v, pcol = q.popleft()
  # 親と異なる色で塗る。
  if colors[v] == -1:
    col = not pcol
    colors[v] = col
    for i in e[v]:
      if colors[i] == -1:
        q.append((i, col))

m = [deque([i for i in range(1, n + 1) if i % 3 == j]) for j in range(3)]

# print(m)

def opop(n):
  if m[n]:
    p = m[n].pop()
  else:
    if n == 1 or n == 2:
      if m[0]:
        p = m[0].pop()
      else:
        p = m[3-n].pop()
    else:
      if m[1]:
        p = m[1].pop()
      else:
        p = m[2].pop()
  return p

ans = [0] * n

if sum(colors) < n // 3:
  for itr, i in enumerate(colors):
    if i:
      ans[itr] = opop(0)
    else:
      ans[itr] = opop(1)
elif sum(colors) > n - (n // 3) + 1:
  for itr, i in enumerate(colors):
    if not i:
      ans[itr] = opop(0)
    else:
      ans[itr] = opop(1)
else:
  for itr, i in enumerate(colors):
    if i:
      ans[itr] = opop(1)
    else:
      ans[itr] = opop(2)
print(*ans)