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)

import sys
from collections import deque

def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LMI(): return list(map(int, sys.stdin.readline().split()))
MOD = 10 ** 9 + 7


N = I()
nl = [list() for _ in range(N)]
for _ in range(N - 1):
    a, b = MI()
    nl[a - 1].append(b - 1)
    nl[b - 1].append(a - 1)

g = {
    0: list(),
    1: list(),
    2: list()
}
q = deque()
q.append((0, -1, 0))
while len(q) > 0:
    v, p, d = q.popleft()
    g[d % 3].append(v)
    for next_v in nl[v]:
        if next_v != p:
            q.append((next_v, v, d + 1))

elnum = list(map(len, g.values()))
elnum.sort()
l, m = divmod(N, 3)
if (m == 0 and [l, l, l] != elnum) or (m == 1 and [l, l, l + 1] != elnum) or (m == 2 and [l, l + 1, l + 1] != elnum):
    print(-1)
    exit(0)


ans = [0] * N
g = list(g.values())
g.sort(key=lambda e: len(e))
for i, v in enumerate(g[0]):
    ans[v] = 3 * (i + 1)
for i, v in enumerate(g[1]):
    ans[v] = 3 * i + 2
for i, v in enumerate(g[2]):
    ans[v] = 3 * i + 1
print(' '.join([str(x) for x in ans]))