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())
lst = [[] for i in range(n+1)]
for i in range(n-1):
    ai, bi = map(int, input().split())
    lst[ai].append(bi)
    lst[bi].append(ai)

tf = [0 for i in range(n+1)]
tf[1] = 1
iro = 1
queue = lst[1]
while queue != []:
    iro = -iro
    new_queue = []
    for i in queue:
        if tf[i] == 0:
            tf[i] = iro
            for j in lst[i]:
                if tf[j] == 0:
                    new_queue.append(j)
    queue = new_queue

n_minus = tf.count(-1)
n_plus = tf.count(1)
ans = [0 for i in range(n)]

if n_minus <= n//3:
    bai = 1
    for i in range(1, n+1):
        if tf[i] == -1:
            ans[i-1] = 3 * bai
            bai += 1
    kouho = 1
    for i in range(1, n+1):
        if tf[i] == 1:
            if bai <= n//3:
                ans[i-1] = 3 * bai
                bai += 1
            else:
                ans[i-1] = kouho
                kouho += 1
                if kouho % 3 == 0:
                    kouho += 1

elif n_plus <= n//3:
    bai = 1
    for i in range(1, n+1):
        if tf[i] == 1:
            ans[i-1] = 3 * bai
            bai += 1
    kouho = 1
    for i in range(1, n+1):
        if tf[i] == -1:
            if bai <= n//3:
                ans[i-1] = 3 * bai
                bai += 1
            else:
                ans[i-1] = kouho
                kouho += 1
                if kouho % 3 == 0:
                    kouho += 1

else:
    bai = 1
    kouho1 = 1
    kouho2 = 2
    for i in range(1, n+1):
        if tf[i] == 1:
            if kouho1 <= n:
                ans[i-1] = kouho1
                kouho1 += 3
            else:
                ans[i-1] = 3 * bai
                bai += 1
        else:
            if kouho2 <= n:
                ans[i-1] = kouho2
                kouho2 += 3
            else:
                ans[i-1] = 3 * bai
                bai += 1

print(' '.join(map(str, ans)))