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: Accepted)

from collections import deque
n=int(input())
graph=[[] for i in range(n)]
for i in range(n-1):
    a,b=map(int,input().split())
    a-=1;b-=1
    graph[a].append(b)
    graph[b].append(a)
even=[]
odd=[]
ansl=[0]*n
nums=set([i for i in range(1,n+1)])
dist=[-1]*n
dist[0]=0
even.append(0)
D=deque([0])
while D:
    v=D.popleft()
    for i in graph[v]:
        if dist[i]!=-1:continue
        dist[i]=dist[v]+1
        D.append(i)
        if dist[i]%2:odd.append(i)
        else:even.append(i)
El=len(even)
Ol=len(odd)
if El <= Ol:
    if El <= n//3:
        for i in range(El):
            ansl[even[i]]=(i+1)*3
            nums.remove((i+1)*3)
        for i in range(Ol):
            ansl[odd[i]]=nums.pop()
    else:
        cnt=0
        while 3*cnt+2<=n:
            ansl[even[cnt]]=3*cnt+2
            ansl[odd[cnt]]=3*cnt+1
            nums.remove(3*cnt+2)
            nums.remove(3*cnt+1)
            cnt+=1
        for i in range(cnt,El):
            ansl[even[i]]=nums.pop()
        for i in range(cnt,Ol):
            ansl[odd[i]]=nums.pop()

if Ol < El:
    if Ol <= n//3:
        for i in range(Ol):
            ansl[odd[i]]=(i+1)*3
            nums.remove((i+1)*3)
        for i in range(El):
            ansl[even[i]]=nums.pop()
    else:
        cnt=0
        while 3*cnt+2<=n:
            ansl[odd[cnt]]=3*cnt+2
            ansl[even[cnt]]=3*cnt+1
            nums.remove(3*cnt+2)
            nums.remove(3*cnt+1)
            cnt+=1
        for i in range(cnt,Ol):
            ansl[odd[i]]=nums.pop()
        for i in range(cnt,El):
            ansl[even[i]]=nums.pop()
print(*ansl)