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:
g = {}
n=int(input())
m = [-1]*n
import sys
sys.setrecursionlimit(1000000000)
def dfs(graph,node,height):
    global m
    global zeros
    global ones
    m[node-1]=height
    if height==0:
      zeros+=1
    if height==1:
      ones+=1
    for n in graph[node]:
      if m[n-1]==-1:
        dfs(graph,n,1-height)

zeros=0
ones=0
for i in range(n-1):
  a,b=map(int,input().split())
  if not a in g:
    g[a]=[b]
  else:
    g[a].append(b)
  if not b in g:
    g[b]=[a]
  else:
    g[b].append(a)
dfs(g,1,0)
ans=[0]*n
used=[0]*n
r=n//3
if 3*(r+1)<=n:
  r+=1
if zeros<=r:
  current=3
  for i in range(n):
    if m[i]==0:
      ans[i]=current
      used[current-1]=1
      current+=3
  counter=0
  for i in range(n):
    if m[i]==1:
      while True:
        if used[counter]==0:
          ans[i]=counter+1
          counter+=1
          break
        counter+=1
elif ones<=r:
  current=3
  for i in range(n):
    if m[i]==1:
      ans[i]=current
      used[current-1]=1
      current+=3
  counter=0
  for i in range(n):
    if m[i]==0:
      while True:
        if used[counter]==0:
          ans[i]=counter+1
          counter+=1
          break
        counter+=1
else:
  if zeros<ones:
    current=1
    for i in range(n):
      if m[i]==0:
        ans[i]=current
        used[current-1]=1
        current+=3
        if current>n:
          current=3
    counter=0
    for i in range(n):
      if m[i]==1:
        while True:
          if used[counter]==0:
            ans[i]=counter+1
            counter+=1
            break
          counter+=1
  else:
    current=1
    for i in range(n):
      if m[i]==1:
        ans[i]=current
        used[current-1]=1
        current+=3
        if current>n:
          current=3
      counter=0
    for i in range(n):
      if m[i]==0:
        while True:
          if used[counter]==0:
            ans[i]=counter+1
            counter+=1
            break
          counter+=1
for i in ans:
  print(i)