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)

import sys
sys.setrecursionlimit(10**6)
def input():
	return sys.stdin.buffer.readline()[:-1]

n = int(input())
adj = [[] for _ in range(n)]
#edges = []
for _ in range(n-1):
	a, b = map(int, input().split())
	#edges.append((a-1, b-1))
	adj[a-1].append(b-1)
	adj[b-1].append(a-1)

if n <= 3:
	print(*list(range(1, n+1)))
	sys.exit()

ans = [-1 for _ in range(n)]
ONE, TWO = n//3 + (n%3 > 0), n//3 + (n%3 > 1)

#parity = [0, 0]
signs = [-1 for _ in range(n)]
def dfs(x):
	for v in adj[x]:
		if signs[v] < 0:
			signs[v] = 1 ^ signs[x]
			dfs(v)

signs[0] = 0
dfs(0)
num1 = sum(signs)
num2 = n - num1

#print(signs)

M, m = max(num1, num2), min(num1, num2) 
if max(num1, num2) > n - (n // 3):
	print(-1)
	sys.exit()
else:
	if num1 > num2:
		one, two = num1, num2
		more = 1
	else:
		one, two = num2, num1
		more = 0
	cur = [2, 1, 3]
	for i in range(n):
		if signs[i] == more:
			if cur[1] > n:
				ans[i] = cur[2]
				cur[2] += 3
			else:
				ans[i] = cur[1]
				cur[1] += 3
		else:
			if cur[0] > n:
				ans[i] = cur[2]
				cur[2] += 3
			else:
				ans[i] = cur[0]
				cur[0] += 3
	print(*ans)