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
n=int(input())
l=[list(map(int,input().split())) for i in range(n-1)]
connection=[[] for i in range(n)]
for i in range(n-1):
  connection[l[i][0]-1].append(l[i][1]-1)
  connection[l[i][1]-1].append(l[i][0]-1)

def bfs(v):
  distance=[-1]*n
  distance[v]=0
  next=connection[v]
  next2=set()
  visited=[-1]*n
  visited[v]=1
  visitct=1
  ct=0
  while len(next)!=0 and visitct!=n:
    ct+=1
    for i in range(len(next)):
      if visited[next[i]]==-1:
        distance[next[i]]=ct
        visited[next[i]]=1
        visitct+=1
        for j in range(len(connection[next[i]])):
          if visited[connection[next[i]][j]]==-1:
            next2.add(connection[next[i]][j])
    next=list(next2)
    next2=set()
  return distance

b=bfs(0)
ct0=0;ct1=0
for i in range(n):
  b[i]%=2
  if b[i]==0:
    ct0+=1
  else:
    ct1+=1

ans=[0]*n
x1=1;x2=2;x3=3

if ct0>=n//3+1 and ct1>=n//3+1:
  for i in range(n):
    if b[i]==0:
      if x1<=n:
        ans[i]=x1
        x1+=3
      else:
        ans[i]=x3
        x3+=3
    else:
      if x2<=n:
        ans[i]=x2
        x2+=3
      else:
        ans[i]=x3
        x3+=3

elif ct0<=n//3:
  for i in range(n):
    if b[i]==0:
      ans[i]=x3
      x3+=3
    else:
      if x1<=n:
        ans[i]=x1
        x1+=3
      elif x2<=n:
        ans[i]=x2
        x2+=3
      else:
        ans[i]=x3
        x3+=3

elif ct1<=n//3:
  for i in range(n):
    if b[i]==1:
      ans[i]=x3
      x3+=3
    else:
      if x1<=n:
        ans[i]=x1
        x1+=3
      elif x2<=n:
        ans[i]=x2
        x2+=3
      else:
        ans[i]=x3
        x3+=3

for i in range(n):
  ans[i]=str(ans[i])
print(' '.join(ans))