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:
from collections import deque


def main():
    N = int(input())
    adj = [set() for _ in range(N)]
    for i in range(N-1):
        a, b = map(int, input().split())
        adj[a-1].add(b-1)
        adj[b-1].add(a-1)
    group = [set() for _ in range(2)]
    q = deque([(0, 0, -1)])
    while len(q):
        cur, g, prev = q.popleft()
        group[g].add(cur)
        for nxt in adj[cur]:
            if nxt == prev:
                continue
            q.append((nxt, 1-g, cur))
    if len(group[0]) > len(group[1]):
        tmp = group[0]
        group[0] = group[1]
        group[1] = tmp
    ans = [0] * N
    pool = set(range(1, N+1))
    if len(group[0]) <= N//3:
        for i, x in enumerate(group[0]):
            ans[x] = (i+1)*3
            pool.remove((i+1)*3)
        for x in group[1]:
            ans[x] = pool.pop()
    else:
        cand = (list(range(1, N+1, 3)) +
                list(range(3, N+1, 3)))[:len(group[0])]
        for x, v in zip(group[0], cand):
            ans[x] = v
            pool.remove(v)
        for x in group[1]:
            ans[x] = pool.pop()
    print(*ans)


if __name__ == "__main__":
    main()