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: Runtime Error)

import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

n = int(input())
G = [[] for i in range(n)]
for i in range(n-1):
    a,b = map(int, input().split())
    a,b = a-1,b-1
    G[a].append(b)
    G[b].append(a)

from collections import deque

def calc_dist(G, start):
    que = deque()
    dep = [-1]*n
    que.append((-1,start,0))
    while que:
        p,v,d = que.pop()
        dep[v]=d
        for to in G[v]:
            if to != p:
                que.append((v,to,d+1))
    return dep

dep0 = calc_dist(G,0)
m = max(dep0)
left = dep0.index(m)
dep1 = calc_dist(G,left)
m = max(dep1)
right = dep1.index(m)
dep2 = calc_dist(G, right)

x = sum(i%2==0 for i in dep1)
y = n - x
ok = set()
for i in range(n):
    ma = max(dep1[i], dep2[i])
    if ma < 3:
        if dep1[i]%2 == 0:
            x -= 1
        else:
            y -= 1
        ok.add(i)

one = []
two = []
three = []
for i in range(1, n+1):
    if i%3 == 0:
        three.append(i)
    elif i%3 == 1:
        one.append(i)
    else:
        two.append(i)

ans = [0]*n
if x<y:
    one, two = two, one
for i in range(n):
    if i in ok:
        continue
    if dep1[i]%2 == 0:
        if one:
            ans[i] = one.pop()
        else:
            ans[i] = three.pop()
    else:
        if two:
            ans[i] = two.pop()
        else:
            ans[i] = three.pop()
res = one+two+three
for i in ok:
    ans[i] = res.pop()

print(*ans)