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:
#!/usr/bin/env python3
import sys, math, itertools, collections, bisect
input = lambda: sys.stdin.buffer.readline().rstrip().decode('utf-8')
inf = float('inf') ;mod = 10**9+7
mans = inf ;ans = 0 ;count = 0 ;pro = 1
sys.setrecursionlimit(10**7)

n = int(input())
G = [[] for i in range(n)]
for _ in range(n-1):
  a,b = map(int,input().split())
  a -= 1; b-=1
  G[a].append(b)
  G[b].append(a)
ans = [0] * n
color = [[],[]]
def coloring(node,pre,oddeven):
  color[oddeven%2].append(node)
  for vi in G[node]:
    if vi != pre:
      coloring(vi,node,oddeven+1)
  return

coloring(0,-1,0)
A = n//3 + int(bool(n%3))
B = n//3 + int(n%3 == 2)
C = n//3
# A>=Bである
a=b=0; c=1
if len(color[0]) >= A and len(color[1]) >= B:
  for i,ci in enumerate(color[0]):
    if i < A:
      ans[ci] = 3 * i + 1
    else:
      ans[ci] = 3 * c
      c += 1

  for i,ci in enumerate(color[1]):
    if i < B:
      ans[ci] = 3 * i + 2
    else:
      ans[ci] = 3 * c
      c += 1

elif len(color[0]) < A:
  # elseが読まれるとき必ずlen(color[0]) < A
  # Bは条件を満たすはず
  for i ,ci in enumerate(color[0]):
    ans[ci] = 3 * c
    c += 1
  for i ,ci in enumerate(color[1]):
    if a < A:
      ans[ci] = 3 * a + 1
      a += 1
    elif b < B:
      ans[ci] = 3 * b + 2
      b += 1
    else:
      ans[ci] = 3 * c
      c += 1

else:
  # elseが読まれるとき必ずlen(color[0]) < A
  # そんなことなかったwwww
  for i ,ci in enumerate(color[1]):
    ans[ci] = 3 * c
    c += 1
  for i ,ci in enumerate(color[0]):
    if a < A:
      ans[ci] = 3 * a + 1
      a += 1
    elif b < B:
      ans[ci] = 3 * b + 2
      b += 1
    else:
      ans[ci] = 3 * c
      c += 1

print(*ans)