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

def resolve():
    N = int(input())
    tree = [[] for i in range(N)]

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

    d = deque([[0, -1]])
    dist = [0] * N
    while d:
      node, par = d.pop()
      children = tree[node]
      for child in children:
        if child == par:
          continue
        dist[child] = (dist[node] + 1) % 2
        d.appendleft([child, node])

    count0 = dist.count(0)
    count1 = N - count0
    s = [set(), set(), set()]
    for i in range(1, N+1):
      s[i%3].add(i)
    if count0 > N//3:
      if count1 > N//3:
        for i in range(N):
          if dist[i]:
            if s[2]:
              dist[i] = s[2].pop()
            else:
              dist[i] = s[0].pop()
          else:
            if s[1]:
              dist[i] = s[1].pop()
            else:
              dist[i] = s[0].pop()
      else:
        for i in range(N):
          if dist[i]:
            dist[i] = s[0].pop()
          else:
            if s[2]:
              dist[i] = s[2].pop()
            elif s[1]:
              dist[i] = s[1].pop()
            else:
              dist[i] = s[0].pop()
    else:
      for i in range(N):
        if dist[i]:
          if s[2]:
            dist[i] = s[2].pop()
          elif s[1]:
            dist[i] = s[1].pop()
          else:
            dist[i] = s[0].pop()
        else:
          dist[i] = s[0].pop()

    print(*dist)
    
resolve()