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: Accepted)

from collections import deque
n = int(input())
G = [[] for i in range(n)]

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

color = [-1] * n  # 木は二分グラフなので彩色

queue = deque([0])
color[0] = 0
red_blue = [1, 0]  # 色毎に塗った数を記録

while queue:
    now = queue.popleft()
    for to in G[now]:
        if color[to] == -1:
            color[to] = 1 - color[now]
            red_blue[color[to]] += 1
            queue.append(to)

ans = [-1] * n  # 異なる色が塗られたもの同士の数字がmod3で同じにならなければ十分
if 3*red_blue[0] < n:
    one, two, three = 0, 0, 0
    for i in range(n):
        if color[i] == 0:
            ans[i] = 3 * three + 3
            three += 1
        else:
            if 3*one + 1 <= n:
                ans[i] = 3 * one + 1
                one += 1
            elif 3*two+2 <= n:
                ans[i] = 3 * two + 2
                two += 1
            else:
                ans[i] = 3 * three + 3
                three += 1
elif 3*red_blue[1] < n:
    one, two, three = 0, 0, 0
    for i in range(n):
        if color[i] == 1:
            ans[i] = 3 * three + 3
            three += 1
        else:
            if 3 * one + 1 <= n:
                ans[i] = 3 * one + 1
                one += 1
            elif 3 * two + 2 <= n:
                ans[i] = 3 * two + 2
                two += 1
            else:
                ans[i] = 3 * three + 3
                three += 1
else:
    one, two, three = 0, 0, 0
    for i in range(n):
        if color[i] == 0:
            if 3*one+1 <= n:
                ans[i] = 3*one+1
                one += 1
            else:
                ans[i] = 3*three+3
                three += 1
        else:
            if 3*two+2 <= n:
                ans[i] = 3*two+2
                two += 1
            else:
                ans[i] = 3*three+3
                three += 1
                
print(*ans)