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:
N = int(input())
ab = [list(map(int,input().split())) for _ in range(N-1)]
from collections import deque
l = [deque() for _ in range(N+1)]
ans = [0]*N
for a,b in ab:
  l[a].append(b)
  l[b].append(a)
  
r = [0]*(N+1)
i = 0
p = l[1]
q = deque()
flag = [False]*(N+1)
flag[1]=True
while (len(p)!=0)or(len(q)!=0):
  if i%2==0:
    while len(p)!=0:
      j = p.pop()
      if flag[j]:
        continue
      else:
        q.extend(l[j])
        flag[j]=True
        r[j]=1
    i+=1
  else:
    while len(q)!=0:
      j = q.pop()
      if flag[j]:
        continue
      else:
        p.extend(l[j])
        flag[j]=True
    i+=1
a = deque(range(1,N+1,3))
b = deque(range(2,N+1,3))
c = deque(range(3,N+1,3))
x = sum(r)
if x<=N//3:
  for i in range(1,N+1):
    if r[i]==1:
      ans[i-1]=c.pop()
    else:
      if len(a)!=0:
        ans[i-1]=a.pop()
      elif len(b)!=0:
        ans[i-1]=b.pop()
      else:
        ans[i-1]=c.pop()
elif (N-x)<=N//3:
  for i in range(1,N+1):
    if r[i]==0:
      ans[i-1]=c.pop()
    else:
      if len(a)!=0:
        ans[i-1]=a.pop()
      elif len(b)!=0:
        ans[i-1]=b.pop()
      else:
        ans[i-1]=c.pop()
else:
  for i in range(1,N+1):
    if r[i]==1:
      if len(a)!=0:
        ans[i-1]=a.pop()
      elif len(c)!=0:
        ans[i-1]=c.pop()
      else:
        ans[i-1]=b.pop()
    else:
      if len(b)!=0:
        ans[i-1]=b.pop()
      elif len(c)!=0:
        ans[i-1]=c.pop()
      else:
        ans[i-1]=a.pop()
        
print(' '.join(map(str,ans)))