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 defaultdict, deque
readline = sys.stdin.readline

inf = 10 ** 5
n = int(readline())
G = defaultdict(list)
deg = [0] * n
for _ in range(n-1):
    a, b = map(int, readline().split())
    a -= 1
    b -= 1
    G[a].append(b)
    G[b].append(a)

q = deque([0])
depth = [-1] * n
depth[0] = 0
d = 1
while q:
    x = q.popleft()
    d = depth[x] + 1
    for y in G[x]:
        if depth[y] == -1:
            q.append(y)
            depth[y] = d
r0 = list(range(3, n + 1, 3))
r1 = list(range(1, n + 1, 3))
r2 = list(range(2, n + 1, 3))
odd = [i for i in range(n) if depth[i] % 2]
even = [i for i in range(n) if not depth[i] % 2]
if len(even) <= n // 3:
    odd, even = even, odd
if len(odd) <= n // 3:
    r = r2 + r1 + r0
else:
    r = r1 + r0 + r2
ans = [0] * n
for i in odd:
    ans[i] = r.pop()
for i in even:
    ans[i] = r.pop()
# print(f'{even=}{odd=}{num=}')
print(*ans)