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())
E = [[] for _ in range(N)]
for _ in range(N-1):
    a,b = map(int, input().split())
    E[a-1].append(b-1)
    E[b-1].append(a-1)

num = [-1]*N
q = deque()
q.append((0, 0))
num[0] = 0
while q:
    temp = q.popleft()
    u = temp[0]
    n = temp[1]
    for v in E[u]:
        if num[v] >= 0:
            continue
        num[v] = (n+1)%2
        q.append((v, (n+1)%2))

ans = [0]*N
if sum(num) <= N//3 or N-sum(num) <= N//3:
    f = 1
    if N-sum(num) <= N//3:
        f = 0
    s = 1
    t = 3
    for i in range(N):
        if num[i] == f:
            ans[i] = t
            t += 3
    for i in range(N):
        if ans[i] == 0:
            ans[i] = s
            s += 1
            if s%3 == 0 and s < t:
                s += 1
else:
    s = 1
    t = 2
    u = 3
    for i in range(N):
        if num[i] == 0 and s <= N:
            ans[i] = s
            s += 3
        elif num[i] == 1 and t <= N:
            ans[i] = t
            t += 3
        else:
            ans[i] = u
            u += 3

print(*ans)