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

N = int(input())
ab = [list(map(int,input().split())) for _ in range(N-1)]
connection = [[] for _ in range(N+1)]
distance3 = [[] for _ in range(N+1)]
remain = [i for i in range(1,N+1)]
Nesessary = [False for _ in range(N+1)]



for node in ab:
    connection[node[0]].append(node[1])
    connection[node[1]].append(node[0])

conne_num = [[i+1, 0] for i in range(N)]

p = 0
for conne in connection[1:]:
    p += 1
    for current1 in conne:
        for current2 in connection[current1]:
            for current3 in connection[current2]:
                if current3 not in conne:
                    distance3[p].append(current3)
                    conne_num[p-1][1] += 1
                    Nesessary[p] = True

conne_num = sorted(conne_num, key=lambda x:int(x[1]), reverse =True)

answer = [0] * (N+1)

for p, conn in enumerate(distance3):
    if len(conn) > 0:
        for i in remain:
           if i % 3 == 0:
               answer[p] = i
               remain.remove(i)
               for q in conn:
                   distance3[q].remove(p)
               break
        else:
            print(-1)
            sys.exit()

for i in range(1,N+1):
    if answer[i] == 0:
        answer[i] = remain[0]
        del remain[0]


print(" ".join(map(str,answer[1:])))