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
n=int(input())
g=[[] for i in range(n)]
for i in range(n-1):
    a,b=map(int,input().split())
    a-=1
    b-=1
    g[a].append(b)
    g[b].append(a)
v=[-1]*n
v[0]=0
q=deque([0])
while q:
    p=q.popleft()
    for i in g[p]:
        if v[i]==-1:
            v[i]=1-v[p]
            q.append(i)
z=0
for i in v:
    if i==0:
        z+=1
ans=[0]*n
if z<=n//3:
    j=3
    for i in range(n):
        if v[i]==0:
            ans[i]=j
            j+=3
    j=1
    for i in range(n):
        if ans[i]==0:
            ans[i]=j
            j+=1
            if j%3==0 and j//3<=z:
                j+=1
    print(*ans)
elif n-z<=n//3:
    j=3
    for i in range(n):
        if v[i]==1:
            ans[i]=j
            j+=3
    j=1
    for i in range(n):
        if ans[i]==0:
            ans[i]=j
            j+=1
            if j%3==0 and j//3<=n-z:
                j+=1
    print(*ans)
else:
    o=1
    t=2
    tt=3
    for i in range(n):
        if v[i]==0 and o<=n:
            ans[i]=o
            o+=3
        elif v[i]==0:
            ans[i]=tt
            tt+=3
        if v[i]==1 and t<=n:
            ans[i]=t
            t+=3
        elif v[i]==1:
            ans[i]=tt
            tt+=3
    print(*ans)