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: Runtime Error)

from collections import deque, defaultdict
import sys
input = sys.stdin.readline
sys.setrecursionlimit(100000000)
n = int(input())
edges = defaultdict(lambda : defaultdict(lambda: False))
for i in range(n-1):
    a, b = map(int, input().split())
    edges[a][b] = True
    edges[b][a] = True

nums = [deque() for _ in range(3)]
for i in range(1, n+1):
    nums[i%3].append(i)

ans = [0 for _ in range(n)]
def dfs(idx, pre_3, pre_2, pre_1):
    if pre_1 == -1:
        if pre_1 % 3 == 2 and len(edges[1]) > 0:
            out = edges[1].pop()
        elif pre_1 % 3 == 1 and len(edges[2]) > 0:
            out = edges[2].pop()
        else:
            out = edges[0].pop()
    else:
        if pre_1 % 3 != 0:
            if pre_1 % 3 == 2 and len(edges[1]) > 0:
                out = nums[1].pop()
            elif pre_1 % 3 == 1 and len(edges[2]) > 0:
                out = nums[2].pop()
            else:
                if len(edges[0]) > 0:
                    out = edges[0].pop()
                else:
                    print(-1)
                    exit()
        else:
            if len(edges[1]) > 0:
                out = nums[1].pop()
            elif len(edges[2]) > 0:
                out = nums[2].pop()
            else:
                if len(edges[0]) > 0:
                    out = edges[0].pop()
                else:
                    print(-1)
                    exit()
    ans[idx-1] = out
    for k in edges[idx].keys():
        if ans[k-1] == 0:
            dfs(step+1, k, pre_2, pre_1, out)

for i in range(1, n+1):
    if len(edges[i]) == 1:
        dfs(1, i, -1, -1, -1)
        break

print(" ".join(list(map(str, ans))))