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


ans = [-1]*(n+1)
q = [1]
ans[1] = 0
while q:
    now = q.pop()
    for next in tree[now]:
        if ans[next] == -1:
            q.append(next)
            ans[next] = (ans[now]+1)%2

if ans.count(1) <= n//3:
    three = set([3*i+3 for i in range(n//3)])
    onetwo = set([i+1 for i in range(n)])
    onetwo = onetwo - three
    three = list(three)
    three.sort()
    onetwo = list(onetwo)
    onetwo.sort()
    three_ind = 0
    onetwo_ind = 0
    for i in range(1, n+1):
        if ans[i] == 1:
            ans[i] = three[three_ind]
            three_ind += 1
        else:
            if onetwo_ind >= len(onetwo):
                ans[i] = three[three_ind]
                three_ind += 1
            else:
                ans[i] = onetwo[onetwo_ind]
                onetwo_ind += 1

elif ans.count(0) <= n//3:
    three = set([3*i+3 for i in range(n//3)])
    onetwo = set([i+1 for i in range(n)])
    onetwo = onetwo - three
    three = list(three)
    three.sort()
    onetwo = list(onetwo)
    onetwo.sort()
    three_ind = 0
    onetwo_ind = 0
    for i in range(1, n+1):
        if ans[i] == 0:
            ans[i] = three[three_ind]
            three_ind += 1
        else:
            if onetwo_ind >= len(onetwo):
                ans[i] = three[three_ind]
                three_ind += 1
            else:
                ans[i] = onetwo[onetwo_ind]
                onetwo_ind += 1

else:
    zero = 0
    one = 0
    two = 0
    for i in range(1, n+1):
        if ans[i] == 0:
            ans[i] = zero*3+1
            zero += 1
        elif ans[i] == 1:
            ans[i] = one*3+2
            one += 1

    three = 0
    for i in range(1, n+1):
        if ans[i] > n:
            ans[i] = three*3+3
            three += 1

print(" ".join(map(str, ans[1:])))