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
import math
mod=10**9+7
inf=float("inf")
from math import sqrt
from collections import deque
from collections import Counter
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from functools import lru_cache
#メモ化再帰defの冒頭に毎回 @lru_cache(maxsize=10**10)
#引数にlistはだめ
#######ここまでテンプレ#######
#######ここから天ぷら########
n=int(input())
G=[[] for i in range(n)]
for i in range(n-1):
    a,b=map(int,input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)

color=[-1 for i in range(n)]
color[0]=0
def coloring(x):
    for i in G[x]:
        if color[i] != -1:continue
        color[i]=1-color[x]
        coloring(i)
coloring(0)
A=[];B=[]
ans=[0 for i in range(n)]
for i in range(n):
    if color[i]==0:A.append(i)
    else:B.append(i)


if 1:
    if len(A)>len(B):
        A,B=B,A
    
    if len(A)<=n//3:
      qq=deque([i for i in range(3,n+1,3)]+[i for i in range(1,n+1,3)]+[i for i in range(2,n+1,3)])
      for i in A:
          ans[i]=qq.popleft()
      for i in B:
          ans[i]=qq.popleft()
      print(*ans)
      exit()
    
    else:
      qq=deque([i for i in range(1,n+1,3)]+[i for i in range(3,n+1,3)]+[i for i in range(2,n+1,3)])
      for i in A:
          ans[i]=qq.popleft()
      for i in B:
          ans[i]=qq.popleft()
      print(*ans)
      exit()