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:
# coding: utf-8
# Your code here!

import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
read = sys.stdin.read


n,*ab = [int(i)-1 for i in read().split()]
n += 1
mp = iter(ab)

g = [[] for _ in range(n)]
for a,b in zip(mp,mp):
    g[a].append(b)
    g[b].append(a)

q = [0]
parent = [-1]*n
depth = [0]*n
#order = []

while q:
    v = q.pop()
    d = depth[v]
    #order.append(v)
    for c in g[v]:
        if c != parent[v]:
            q.append(c)
            parent[c] = v
            depth[c] = d+1

odd = [i for i,d in enumerate(depth) if d&1]
even = [i for i,d in enumerate(depth) if d&1==0]

s0 = list(range(3,n+1,3))
s1 = list(range(1,n+1,3))
s2 = list(range(2,n+1,3))

ans = [0]*n

m = n//3
if len(even) <= m:
    odd,even = even,odd

if len(odd) <= m:
    s2 += s1+s0
else:
    s2 += s0+s1

for i in odd:
    ans[i] = s2.pop()
for i in even:
    ans[i] = s2.pop()
    

print(*ans)