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:
import sys
input = sys.stdin.readline

n = int(input())

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

used = [0]*n
tank = [0]
used[0] = 1
ca = 0
cb = 0
while tank:
    p = tank.pop()
    if used[p] > 0:
        ca += 1
    for e in edge[p]:
        if used[e] == 0:
            used[e] = -used[p]
            tank.append(e)
cb = n-ca
res = [0]*n
flag = [True]*(n+2)

if ca <= n//3:
    tmp = 0
    for i in range(n):
        if used[i] == 1:
            tmp += 3
            res[i] = tmp
            flag[tmp] = False
    tmp = 1
    for i in range(n):
        if used[i] != 1:
            res[i] = tmp
            flag[tmp] = False
            while not flag[tmp]:
                tmp += 1
    print(*res)
elif cb <= n//3:
    tmp = 0
    for i in range(n):
        if used[i] == -1:
            tmp += 3
            res[i] = tmp
            flag[tmp] = False
    tmp = 1
    for i in range(n):
        if used[i] != -1:
            res[i] = tmp
            flag[tmp] = False
            while not flag[tmp]:
                tmp += 1
    print(*res)
else:
    tmp = -2
    for i in range(n):
        if used[i] == -1:
            tmp += 3
            if tmp > n:
                tmp = 3
            res[i] = tmp
            flag[tmp] = False
    tmp = 2
    for i in range(n):
        if used[i] != -1:
            res[i] = tmp
            flag[tmp] = False
            while not flag[tmp]:
                tmp += 1
                if tmp > n:
                    tmp = 3
                    break
    print(*res)