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**9)
N=int(input())

tree=[[] for _ in range(N+1)]
for _ in range(N-1):
  a,b=map(int,input().split())
  tree[a].append(b)
  tree[b].append(a)
#print(tree)

def dfs(u,par,depth):
  dist_list[u]=depth
  for v in tree[u]:
    if v!=par:
      dfs(v,u,depth+1)
  
dist_list=[-1]*(N+1)
dfs(1,0,0)
#print(dist_list)

num_even=num_odd=0
for i in range(1,N+1):
  if dist_list[i]%2==0:
    num_even+=1
  else:
    num_odd+=1
#print(num_even,num_odd)

num_3m0=3
num_3m1=1
num_3m2=2
answer_list=[0]*(N+1)
if num_even<=N//3:
  for i in range(1,N+1):
    if dist_list[i]%2==0:
      answer_list[i]=num_3m0
      num_3m0+=3
    else:
      if num_3m1<=N:
        answer_list[i]=num_3m1
        num_3m1+=3
      elif num_3m2<=N:
        answer_list[i]=num_3m2
        num_3m2+=3
      else:
        answer_list[i]=num_3m0
        num_3m0+=3
elif num_odd<=N//3:
  for i in range(1,N+1):
    if dist_list[i]%2==1:
      answer_list[i]=num_3m0
      num_3m0+=3
    else:
      if num_3m1<=N:
        answer_list[i]=num_3m1
        num_3m1+=3
      elif num_3m2<=N:
        answer_list[i]=num_3m2
        num_3m2+=3
      else:
        answer_list[i]=num_3m0
        num_3m0+=3
else:
  for i in range(1,N+1):
    if dist_list[i]%2==0:
      if num_3m1<=N:
        answer_list[i]=num_3m1
        num_3m1+=3
      else:
        answer_list[i]=num_3m0
        num_3m0+=3
    else:
      if num_3m2<=N:
        answer_list[i]=num_3m2
        num_3m2+=3
      else:
        answer_list[i]=num_3m0
        num_3m0+=3
        
print(*answer_list[1:])