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,*t=map(int,open(0).read().split())
e=[[]for _ in range(n)]
x=[[],[],[]]
for i in range(n):
    x[i%3]+=i+1,
for a,b in zip(*[iter(t)]*2):
    e[a-1]+=b-1,
    e[b-1]+=a-1,
a=[-1]*n
p=a[:]
p[0]=0
s=[0]
c=[1,0]
while s:
    v=s.pop()
    for w in e[v]:
        if p[w]<0:
            p[w]=p[v]^1
            c[p[w]]+=1
            s+=w,
if min(c)<=n//3:
    x[0]+=x[1]
    f=c[0]<=n//3
    for i,v in enumerate(p):
        j=(v==f)-1
        a[i]=x[j if x[j]else-1].pop()
else:
    for i,v in enumerate(p):
        a[i]=x[v if x[v]else-1].pop()
print(*a)