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
sys.setrecursionlimit(10**7)

n = int(input())
T = [[] for _ in range(n)]
for _ in range(n-1):
  a, b = map(int, input().split())
  a -= 1
  b -= 1
  T[a].append(b)
  T[b].append(a)
L = [-1]*n
dist = [-1]*n
dist[0] = 0
F_1 = [i for i in range(1, n+1) if i%3==1]
F_2 = [i for i in range(1, n+1) if i%3==2]
F_0 = [i for i in range(1, n+1) if i%3==0]
def dfs(v):
  for nv in T[v]:
    if dist[nv] == -1:
      dist[nv] = dist[v] ^ 1
      dfs(nv)
dfs(0)
l0 = dist.count(0)
l1 = n - l0
if l0 <= n//3:
  for i in range(n):
    if dist[i] == 0:
      L[i] = F_0.pop()
    else:
      if F_1:
        L[i] = F_1.pop()
      elif F_2:
        L[i] = F_2.pop()
      else:
        L[i] = F_0.pop()
elif l1 <= n//3:
  for i in range(n):
    if dist[i]:
      L[i] = F_0.pop()
    else:
      if F_1:
        L[i] = F_1.pop()
      elif F_2:
        L[i] = F_2.pop()
      else:
        L[i] = F_0.pop()
else:
  for i in range(n):
    if dist[i]:
      if F_1:
        L[i] = F_1.pop()
      else:
        L[i] = F_0.pop()
    else:
      if F_2:
        L[i] = F_2.pop()
      else:
        L[i] = F_0.pop()
print(*L)