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
input = sys.stdin.readline
sys.setrecursionlimit(10**5)

N = int(input())
vec = [[] for _ in range(N)]
for _ in range(N - 1) :
  a, b = map(int, input().split())
  vec[a-1].append(b-1)
  vec[b-1].append(a-1)
  
q = [(0, -1, 0)]
dist = [0] * N
cnt = [0, 0]
while q :
  cur, pre, cost = q.pop()
  dist[cur] = cost
  cnt[cost] += 1
  for nex in vec[cur] :
    if nex != pre :
      q.append((nex, cur, cost ^ 1))

one = 1
two = 2
three = 3
if cnt[0] < (N + 2) // 3 :
  for i in range(N) :
    if dist[i] == 0 :
      dist[i] = three
      three += 3
    elif one <= N :
      dist[i] = one
      one += 3
    elif two <= N :
      dist[i] = two
      two += 3
    else :
      dist[i] = three
      three += 3
      
elif cnt[1] < (N + 1) // 3 :
  for i in range(N) :
    if dist[i] == 1 :
      dist[i] = three
      three += 3
    elif one <= N :
      dist[i] = one
      one += 3
    elif two <= N :
      dist[i] = two
      two += 3
    else :
      dist[i] = three
      three += 3
      
else :
  for i in range(N) :
    if dist[i] == 0 and one <= N :
      dist[i] = one
      one += 3
    elif dist[i] == 1 and two <= N :
      dist[i] = two
      two += 3
    else :
      dist[i] = three
      three += 3
      
print(*dist)