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

import sys
input = sys.stdin.buffer.readline

N = int(input())
edges = [[] for _ in range(N)]

for _ in range(N - 1):
    fr, to = map(lambda a: int(a) - 1, input().split())
    edges[fr].append(to)
    edges[to].append(fr)

dist = [10**10] * N
que = deque([(0, 0)])
while que:
    now, d = que.popleft()

    if dist[now] <= d:
        continue
    dist[now] = d

    for to in edges[now]:
        que.append((to, d + 1))

A = [i for i, d in enumerate(dist) if d % 2 == 0]
B = [i for i, d in enumerate(dist) if d % 2 == 1]

if len(A) > len(B):
    A, B = B, A

ans = [-1] * N
nums = set(range(1, N + 1))

if len(A) <= N // 3:
    mul = 1
    for i in A:
        ans[i] = 3 * mul
        nums.remove(3 * mul)
        mul += 1
    nums = list(nums)
    for i, n in zip(B, nums):
        ans[i] = n
else:
    mul = 1
    for c, i in enumerate(A):
        if c * 3 + 1 > N:
            ans[i] = mul * 3
            mul += 1
        else:
            ans[i] = c * 3 + 1
    for c, i in enumerate(B):
        if c * 3 + 2 > N:
            ans[i] = mul * 3
            mul += 1
        else:
            ans[i] = c * 3 + 2

print(*ans)