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)

from collections import deque
from itertools import permutations

def BFS(graph, start, end):
    search_dict = {}
    mark_dict = {}
    queue = deque([start])
    mark_dict[start] = 0
    while len(queue) > 0:
        target = queue.pop()
        if mark_dict[target] > 3:
            break
        if target not in search_dict:
            if target != end:
                for elem in graph[target]:
                    search_dict[target] = True
                    mark_dict[elem] = mark_dict[target] + 1
                    queue.appendleft(elem)
            else:
                return mark_dict[target] == 3
    return False

n = int(input())
graph = {}
for i in range(n-1):
    a, b = map(str, input().split())
    if not(a in graph):
        graph[a] = []
    if not(b in graph):
        graph[b] = []
    graph[a].append(b)
    graph[b].append(a)

result = []
calc = [[False] * (n + 1) for i in range(n + 1)]
for i in range(n):
    for j in range(n - i):
        newi = i + 1
        newj = j + 1
        if (newi + newj) % 3 == 0 or newi * newj % 3 == 0:
            calc[newj][newi] = True
            calc[newi][newj] = True
        if newi == newj + i:
            continue
        if BFS(graph, str(newi), str(newj + i)):
            result.append([newi, newj + i])


for per in list(permutations(range(n))):
    ok = True
    for test in result:
        if calc[per[int(test[0]) - 1]][per[int(test[1]) - 1]]:
            continue
        ok = False
    if ok:
        print(per)

print("-1")