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: Wrong Answer)

def dijkstra(s, n, w, cost):
    # 始点sから各頂点への最短距離
    # n:頂点数,　w:辺の数, cost[u][v] : 辺uvのコスト(存在しないときはinf)
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0

    while True:
        v = -1
        # まだ使われてない頂点の中から最小の距離のものを探す
        for i in range(n):
            if (not used[i]) and (v == -1):
                v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
            break
        used[v] = True

        for j in range(n):
            d[j] = min(d[j], d[v] + cost[v][j])
    return d

################################
#n, w = map(int, input().split())  # n:頂点数　w:辺の数
n = int(input())  # n:頂点数　w:辺の数
w = n-1

cost = [[float("inf") for i in range(n)] for i in range(n)]
# cost[u][v] : 辺uvのコスト(存在しないときはinf この場合は10**10)
z=1
for i in range(w):
    # x, y, z = map(int, input().split())
    x, y = map(int, input().split())
    cost[x-1][y-1] = z
    cost[y-1][x-1] = z
#print(dijkstra(1, n, w, cost))
li1 = []
for i in range(n):
    for j in range(n):
        k = dijkstra(i, n, w, cost)
        if k[j]==3:
            li1.append([i,j])
ans=[1]*n
for i in li1:
    ans[i[0]]=3
    ans[i[1]]=3

if li1 == []:
    print('-1')
else:
    print(ans)