Problem: 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.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in PYTHON3):

Solution:
def main():
    from collections import deque
    from itertools import chain
    import sys
    input = sys.stdin.readline

    N = int(input())

    graph = tuple(set() for _ in range(N))
    for _ in range(N - 1):
        a, b = (int(x) - 1 for x in input().split())
        graph[a].add(b)
        graph[b].add(a)

    bipartite = [-1] * N
    bipartite[0] = 0

    dq = deque([0])
    while dq:
        v = dq.popleft()
        nb = bipartite[v] ^ 1
        for u in graph[v]:
            if ~bipartite[u]: continue
            bipartite[u] = nb
            dq.append(u)

    s0 = bipartite.count(0)
    s1 = N - s0

    it1 = iter(range(1, N + 1, 3))
    it2 = iter(range(2, N + 1, 3))
    it3 = iter(range(3, N + 1, 3))
    it123 = chain(it1, it2, it3)
    it13 = chain(it1, it3)
    it23 = chain(it2, it3)

    ans = [-1] * N
    K = N // 3
    if s0 <= K:
        for v in range(N):
            if bipartite[v] == 0:
                ans[v] = next(it3)
            else:
                ans[v] = next(it123)  # 余った3をこちらで消費するので12ではなく123
    elif s1 <= K:
        for v in range(N):
            if bipartite[v] == 1:
                ans[v] = next(it3)
            else:
                ans[v] = next(it123)
    else:
        for v in range(N):
            if bipartite[v] == 0:
                ans[v] = next(it13)
            else:
                ans[v] = next(it23)

    print(*ans)


if __name__ == '__main__':
    main()