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,*L=map(int,open(0).read().split())
g=[[]for _ in range(n)]
for a,b in zip(*[iter(L)]*2):
	g[a-1].append(b-1)
	g[b-1].append(a-1)
tr=[-1]*n
q=[0]
tr[0]=0
while q:
	cu=q.pop()
	for nx in g[cu]:
		if tr[nx]<0:
			tr[nx]=1-tr[cu]
			q.append(nx)
lab=[-1]*n
z,o=map(tr.count,(0,1))
if z<=n//3 or o<=n//3:
	f=0 if z<=n//3 else 1
	onetwo=1
	thr=3
	for i,x in enumerate(tr):
		if x==f:
			lab[i]=thr
			thr+=3
	for i,x in enumerate(tr):
		if x==1-f:
			if thr<=n:
				lab[i]=thr
				thr+=3
			else:
				lab[i]=onetwo
				onetwo+=1 if onetwo%3==1 else 2
else:
	one=1
	for i,x in enumerate(tr):
		if x==0:
			lab[i]=one
			one+=3
			if one>n:
				break
	two=2
	for i,x in enumerate(tr):
		if x==1:
			lab[i]=two
			two+=3
			if two>n:
				break
	thr=3
	for i in range(n):
		if lab[i]<0:
			lab[i]=thr
			thr+=3
print(*lab)