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 main():
    N = int(input())
    T = [[] for _ in range(N)]
    for _ in range(N - 1):
        a, b = map(int, input().split())
        a, b = a - 1, b - 1
        T[a].append(b)
        T[b].append(a)


    def calc_dis3(s):
        stack = [(s, 1)]
        visited = [0] * N
        dis3 = set()

        while stack:
            n, depth = stack.pop()
            visited[n] = 1

            if depth == 4:
                dis3.add(n)
                continue

            for to in T[n]:
                if visited[to]:
                    continue

                stack.append((to, depth + 1))

        return dis3


    dis3_dict = {i: None for i in range(N)}
    dis3_counts = [0] * N
    for n in range(N):
        ret_calc_dis3 = calc_dis3(n)
        dis3_dict[n] = ret_calc_dis3
        dis3_counts[n] = len(ret_calc_dis3)


    # ３の倍数をできるだけいい位置にセット
    ans = [-1] * N
    dis3_counts_sorted = sorted([[c, i] for i, c in enumerate(dis3_counts)], reverse=True)
    for i in range(N // 3):
        ans[dis3_counts_sorted[i][1]] = 3 * (i + 1)


    # 残ったやつをいい感じにset
    x1 = [n for n in range(1, N + 1) if n % 3 == 1]
    x2 = [n for n in range(1, N + 1) if n % 3 == 2]
    for n in range(N):
        if ans[n] > 0:
            continue

        px = 1 if len(x1) >= len(x2) else 2
        if px == 1:
            ans[n] = x1.pop()
        else:
            ans[n] = x2.pop()

        for to in dis3_dict[n]:
            if (n > to) or (ans[to] > 0):
                continue

            if px == 1:
                if not x2:
                    continue
                ans[to] = x2.pop()
            else:
                if not x1:
                    continue
                ans[to] = x1.pop()

    remains = x1 + x2
    for n in range(N):
        if ans[n] > 0:
            continue
        ans[n] = remains.pop()


    def check(s):
        stack = [(s, 1)]
        visited = [0] * N

        while stack:
            n, depth = stack.pop()
            visited[n] = 1

            if depth == 4:
                continue

            for to in T[n]:
                if visited[to]:
                    continue

                if depth + 1 == 4:
                    if not (((ans[s] * ans[to]) % 3) == 0 or ((ans[s] + ans[to]) % 3 == 0)):
                        print(n, to)
                        print(-1)
                        exit()
                stack.append((to, depth + 1))

    for n in range(N):
        check(n)

    print(*ans, sep=' ')


if __name__ == '__main__':
    main()