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 *
from itertools import *
from bisect import *
from heapq import *
import copy
import math
from fractions import gcd


N=int(input())
AB=[list(map(int,input().split())) for i in range(N-1)]

data=[[] for i in range(N+1)]
for a,b in AB:
    data[a].append(b)
    data[b].append(a)

lst=[0 for i in range(N+1)]
stack=deque([[1,0]])
visited=set([1])

while stack:
    a,m=stack.popleft()
    for p in data[a]:
        if not p in visited:
            visited.add(p)
            stack.append([p,m+1])
            lst[p]=m+1


oddslst=[i for i in range(len(lst)) if i and not lst[i]%2]
evenlst=[i for i in range(len(lst)) if i and lst[i]%2]
lst=[0 for i in range(N+1)]
if N//3<len(oddslst) and N//3<len(evenlst):
    for i in range(1,N+1):
        if i%3==1:
            lst[oddslst.pop()]=i
        if i%3==2:
            lst[evenlst.pop()]=i
    oddslst+=evenlst
    for i in range(1,N+1):
        if i%3==0:
            lst[oddslst.pop()]=i
else:
    if len(evenlst)<len(oddslst):
        oddslst,evenlst=evenlst,oddslst
    for i in range(1,N+1):
        if i%3==0:
            if oddslst:
                lst[oddslst.pop()]=i
            else:
                lst[evenlst.pop()]=i
        else:
            lst[evenlst.pop()]=i

print(" ".join([str(i) for i in lst if i]))