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:
n=int(input())
ki=[[] for _ in range(n)]

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

odd=[-1]*n
odd[0]=0
todo=[0]
while todo:
    v=todo.pop()
    for nv in ki[v]:
        if odd[nv]==-1:
            odd[nv]=abs(odd[v]-1)
            todo.append(nv)
from collections import Counter
co=Counter(odd)
ans=[0]*n
if co[1]<=n//3:
    cnt1=1
    cnt2=0
    cnt3=1
    for i in range(n):
        if odd[i]==1 and 3*cnt3<=n:
            ans[i]=3*cnt3
            cnt3+=1
        else:
            if cnt1%3==0:cnt1+=1
            if cnt1<=n:
                ans[i]=cnt1
                cnt1+=1
            else:
                ans[i]=n-n%3-cnt2*3
                cnt2+=1

elif co[0]<=n//3:
    cnt1=1
    cnt2=0
    cnt3=1
    for i in range(n):
        if odd[i]==0 and 3*cnt3<=n:
            ans[i]=3*cnt3
            cnt3+=1
        else:
            if cnt1%3==0:cnt1+=1
            if cnt1<=n:
                ans[i]=cnt1
                cnt1+=1
            else:
                ans[i]=n-n%3-cnt2*3
                cnt2+=1
else:
    cnt1=1
    cnt2=1
    cnt3=1
    for i in range(n):
        if odd[i]==1:
            if 3*cnt1-2<=n:
                ans[i]=3*cnt1-2
                cnt1+=1
            else:
                ans[i]=3*cnt3
                cnt3+=1
        else:
            if 3*cnt2-1<=n:
                ans[i]=3*cnt2-1
                cnt2+=1
            else:
                ans[i]=3*cnt3
                cnt3+=1
print(*ans,sep=' ')