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
input=lambda: sys.stdin.readline().rstrip()
n=int(input())

edge = [[] for i in range(n)]
for i in range(n-1):
  a,b=map(int,input().split())
  edge[a-1].append(b-1)
  edge[b-1].append(a-1)

inf=float("inf")

prv=[inf]*n
prv[0]=-1
depth=[inf]*n
depth[0]=0
stack=[0]
while stack:
  cur=stack.pop()
  for next in edge[cur]:
    if prv[next]==inf:
      prv[next]=cur
      depth[next]=depth[cur]+1
      stack.append(next)
e,o=0,0
for i in range(n):
  if depth[i]%2:
    o+=1
  else:
    e+=1
Ans=[0]*n
if e<=n//3:
  ct1,ct2,ct3=0,0,0
  for i in range(n):
    if not depth[i]%2:
      Ans[i]=3*ct3+3
      ct3+=1
    else:
      if 3*ct2+2<=n:
        Ans[i]=3*ct2+2
        ct2+=1
      elif 3*ct1+1<=n:
        Ans[i]=3*ct1+1
        ct1+=1
      else:
        Ans[i]=3*ct3+3
        ct3+=1
elif o<=n//3:
  ct1,ct2,ct3=0,0,0
  for i in range(n):
    if depth[i]%2:
      Ans[i]=3*ct3+3
      ct3+=1
    else:
      if 3*ct1+1<=n:
        Ans[i]=3*ct1+1
        ct1+=1
      elif 3*ct2+2<=n:
        Ans[i]=3*ct2+2
        ct2+=1
      else:
        Ans[i]=3*ct3+3
        ct3+=1
else:
  ct1,ct2,ct3=0,0,0
  for i in range(n):
    if not depth[i]%2:
      if 3*ct1+1<=n:
        Ans[i]=3*ct1+1
        ct1+=1
      else:
        Ans[i]=3*ct3+3
        ct3+=1
    else:
      if 3*ct2+2<=n:
        Ans[i]=3*ct2+2
        ct2+=1
      else:
        Ans[i]=3*ct3+3
        ct3+=1

print(*Ans)