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:
from collections import deque
N, *L = map(int, open(0).read().split())
E = [[] for i in range(N+1)]
for a,b  in zip(*[iter(L)]*2):
  E[a].append(b)
  E[b].append(a)
q = deque([1])
dist = [-1]*(N+1)
dist[1] = 0
dic = [[1],[]]
while q:
  u = q.popleft()
  for v in E[u]:
    if dist[v]==-1:
      dist[v] = dist[u]+1
      q.append(v)
      dic[dist[v]%2].append(v)

a = len(dic[0])
b = len(dic[1])
#print(dic)
seq = set(range(1,N+1))
ver = set(range(1,N+1))
ans = [0]*(N+1)
if a>N//3 and b>N//3:
  for i in range(1,N+1,3):
    v = dic[1].pop()
    ans[v] = i
  q = 0
  for i in range(2,N+1,3):
    v = dic[0].pop()
    ans[v] = i
  m = 3
  while dic[0]:
    v = dic[0].pop()
    ans[v] = m
    m += 3
  while dic[1]:
    v = dic[1].pop()
    ans[v] = m
    m += 3
elif a<=N//3:
  seq = set(range(1,N+1))
  m = 3
  while dic[0]:
    v = dic[0].pop()
    ans[v] = m
    seq.remove(m)
    m += 3
  seq = list(seq)
  while dic[1]:
    v = dic[1].pop()
    ans[v] = seq.pop()
else:
  seq = set(range(1,N+1))
  m = 3
  while dic[1]:
    v = dic[1].pop()
    ans[v] = m
    seq.remove(m)
    m += 3
  seq = list(seq)
  while dic[0]:
    v = dic[0].pop()
    ans[v] = seq.pop()
print(*ans[1:])