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 collections
n = int(input())
one = collections.deque([1])
two = collections.deque([])
three = collections.deque([])
graph = [[] for _ in range(n)]

for i in range(n - 1):
    a, b = map(int, input().split())
    graph[a - 1].append(b - 1)
    graph[b - 1].append(a - 1)
    if i % 3 == 0: two.append(i + 2)
    elif i % 3 == 1: three.append(i + 2)
    else: one.append(i + 2)

que = collections.deque([0])
num = [0] * n
dist = [-1] * n
parity = [0] * 2
dist[0] = 0
flag = True
while len(que) != 0:
    i = que.popleft()
    parity[dist[i] % 2] += 1
    for j in graph[i]:
        if dist[j] != -1: continue
        dist[j] = dist[i] + 1
        que.append(j)

if parity[0] <= n // 3 or parity[1] <= n // 3:
    rem = 0 if parity[0] < parity[1] else 1
    for i, v in enumerate(dist):
        if v % 2 == rem:
            num[i] = str(three.popleft())
        else:
            if len(one) != 0: num[i] = str(one.popleft())
            elif len(two) != 0: num[i] = str(two.popleft())
            else: num[i] = str(three.popleft())
else:
    for i, v in enumerate(dist):
        if v % 2 == 0:
            if len(one) != 0: num[i] = str(one.popleft())
            elif len(three) != 0: num[i] = str(three.popleft())
            else:
                flag = False
                break
        else:
            if len(two) != 0: num[i] = str(two.popleft())
            elif len(three) != 0: num[i] = str(three.popleft())
            else:
                flag = False
                break

print(' '.join(num))