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 ** 6)
n = int(input())
 
edge = [[] for i in range(n)]
ans = [-1] * n
odd = []
even =[]
nums = [[],[],[]]
for i in range(1,n+1):
    nums[i % 3].append(i)
 
def dfs(x, last = -1, d = 0):
    if d % 2 == 0: even.append(x)
    else: odd.append(x)
    for nxt in edge[x]:
        if nxt == last:continue
        dfs(nxt, x, d + 1)
 
for i in range(n-1):
    a, b = map(int, input().split())
    a-=1
    b-=1
    edge[a].append(b)
    edge[b].append(a)
 
dfs(0)
 
zerocnt = len(nums[0])
 
if len(even) <= zerocnt:
    for even_node in even:
        ans[even_node] = nums[0].pop()
    rest = nums[0] + nums[1] + nums[2]
    for odd_node in odd:
        ans[odd_node] = rest.pop()
 
elif len(odd) <= zerocnt:
    for odd_node in odd:
        ans[odd_node] = nums[0].pop()
    rest = nums[0] + nums[1] + nums[2]
    for even_node in even:
        ans[even_node] = rest.pop()
 

else:
    for num in nums[1]:
        ans[even.pop()] = num
    for num in nums[2]:
        ans[odd.pop()] = num
    rest = nums[0]
    for even_node in even:
        ans[even_node] = rest.pop()
    for odd_node in odd:
        ans[odd_node] = rest.pop()
 
print(*ans)