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:
import sys
from collections import deque

read = sys.stdin.read
readline = sys.stdin.readline

N, *ab = map(int, read().split())

if N == 2:
    print(1, 2)
    exit()

graph = [[] for _ in range(N + 1)]
for a, b in zip(*[iter(ab)] * 2):
    graph[a].append(b)
    graph[b].append(a)

k0 = list(range(3, N + 1, 3))
k1 = list(range(1, N + 1, 3))  # 1番多い
k2 = list(range(2, N + 1, 3))

start = ab[0]
color = 1  # 1が黒
queue = deque([(start, color)])
colors = [0] * (N + 1)
colors[start] = 1
while queue:
    V, c = queue.popleft()
    c = -c
    for v in graph[V]:
        if colors[v] == 0:
            colors[v] = c
            queue.append((v, c))

black = colors.count(1)
white = N - black
three = (N - 3) // 3 + 1
if black <= three:
    b = k0[:black]
    w = k1 + k2 + k0[black:]
    answer = [b.pop() if i == 1 else w.pop() for i in colors[1:]]
    print(' '.join(map(str, answer)))
elif white <= three:
    w = k0[:white]
    b = k1 + k2 + k0[white:]
    answer = [b.pop() if i == 1 else w.pop() for i in colors[1:]]
    print(' '.join(map(str, answer)))
else:
    answer = [0] * (N + 1)
    for i in range(1, N + 1):
        if colors[i] == 1:
            if k1:
                answer[i] = k1.pop()
            else:
                answer[i] = k0.pop()
        else:
            if k2:
                answer[i] = k2.pop()
            else:
                answer[i] = k0.pop()
    print(' '.join(map(str, answer[1:])))