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 heapq import heappop as he
n=int(input())
s=[[]for i in range(n)]
for i in range(n-1):
    a,b=map(int,input().split())
    a,b=a-1,b-1
    s[a].append(b)
    s[b].append(a)
c=[0]*n
q=[1]
t=[1]*n
t[1]=0
fl=0
while q:
    fl^=1
    p=[]
    for j in q:
        for h in s[j]:
            if t[h]:
                t[h]=0
                c[h]=fl
                p.append(h)
    q=p.copy()
if c.count(1)>n//3 and c.count(0)>n//3:
    p,q,r=1,2,3
    for i in range(n):
        if c[i]:
            if q<=n:
                c[i]=q
                q+=3
            else:
                c[i]=r
                r+=3
        else:
            if p<=n:
                c[i]=p
                p+=3
            else:
                c[i]=r
                r+=3
elif c.count(1)<=n//3:
    p,q,r=1,2,3
    for i in range(n):
        if c[i]:
            c[i]=r
            r+=3
        else:
            if p<=n:
                c[i]=p
                p+=3
            elif q<=n:
                c[i]=q
                q+=3
            else:
                c[i]=r
                r+=3
else:
    p,q,r=1,2,3
    for i in range(n):
        if not c[i]:
            c[i]=r
            r+=3
        else:
            if q<=n:
                c[i]=q
                q+=3
            elif p<=n:
                c[i]=p
                p+=3
            else:
                c[i]=r
                r+=3
print(*c)