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: Runtime Error)

N = int(input())
AB = [list(map(int, input().split())) for _ in range(N - 1)]
# N = 9
# AB = [[1, 2], [2, 3], [9, 8], [8, 3], [3, 4], [4, 5], [3, 6], [6, 7]]
if N == 2:
    print(1, 2)
if N == 3:
    print(1, 2, 3)

x1 = set()
x2 = set()
x3 = set()

for i in range(1, N + 1):
    if i % 3 == 1:
        x1.add(i)
    elif i % 3 == 2:
        x2.add(i)
    else:
        x3.add(i)
x = [x1, x2, x3]

result = {}


class Node:
    def __init__(self, v):
        self.linked = set()
        self.v = v

    def add_node(self, node):
        self.linked.add(node)

    def set_rank(self, parent, rank):
        self.rank = rank
        if parent == None:
            result[self.v] = x1.pop()
        else:
            if x[(result[self.parent.v] + 1) % 3]:
                result[self.v] = x[(result[self.parent.v] + 1) % 3].pop()
            elif x1:
                x1.pop()
            elif x2:
                x2.pop()
            else:
                x3.pop()
        for node in self.linked:
            if node == parent:
                continue
            node.set_rank(parent=self, rank=rank + 1)

    def __repr__(self):
        return '{}'.format(self.rank)


nodes = {v: Node(v) for v in range(1, N + 1)}
for a, b in AB:
    nodes[a].add_node(nodes[b])
    nodes[b].add_node(nodes[a])

# usokaihou
root = nodes[1]
root.set_rank(None, 0)
print(' '.join(str(result[i]) for i in range(1, N + 1)))

exit()