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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
import math
mod=10**9+7
inf=float("inf")
from math import sqrt
from collections import deque
from collections import Counter
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from functools import lru_cache
#メモ化再帰defの冒頭に毎回 @lru_cache(maxsize=10**10)
#引数にlistはだめ
#######ここまでテンプレ#######
#######ここから天ぷら########
n=int(input())
G=[[] for i in range(n)]
for i in range(n-1):
    a,b=map(int,input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)

color=[-1 for i in range(n)]
color[0]=0
def coloring(x):
    for i in G[x]:
        if color[i] != -1:continue
        color[i]=1-color[x]
        coloring(i)
coloring(0)
A=[];B=[]
ans=[0 for i in range(n)]
for i in range(n):
    if color[i]==0:A.append(i)
    else:B.append(i)
if len(A)<=n//3 or len(B)<=n//3:
    if len(A)>len(B):
        A,B=B,A
    qq=deque([3*i+1 for i in range(int(n/3+0.9999))]+[3*i for i in range(int(n//3),0,-1)]+[3*i+2 for i in range(int(n/3+0.9999))])
    for i in A:
        ans[i]=qq.popleft()
        
    
    for t in range(n):
        if ans[t]==0:
            ans[t]=qq.popleft()
            
    print(*ans)
    exit()
qq=deque([3*i+1 for i in range(int(n/3+0.9999))]+[3*i for i in range(int(n//3),0,-1)])
for i in A:
    ans[i]=qq.popleft()
qq.extend([3*i+2 for i in range(int(n/3+0.9999))])
for j in B:
    ans[j]=qq.pop()
print(*ans)