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
from collections import deque
sys.setrecursionlimit(10 ** 6)


def dfs(v, step):
    global node0, node1
    if step & 1:
        color[v] = 1
        node1 += 1
    else:
        color[v] = 0
        node0 += 1
    for nv in edge[v]:
        if color[nv] is None:
            dfs(nv, step + 1)


N = int(input())
edge = [[] for _ in range(N)]
for s in sys.stdin.readlines():
    a, b = map(int, s.split())
    a -= 1; b -= 1
    edge[a].append(b)
    edge[b].append(a)
color = [None] * N
node0 = node1 = 0
dfs(0, 0)

ans = [None] * N
num0 = N // 3
num1 = N // 3 + (N % 3 >= 1)
num2 = N // 3 + (N % 3 == 2)
q0 = deque()
q1 = deque()
q = deque()
if node0 > num0 and node1 > num0:
    for i in range(1, num0 + 1):
        q.append(3 * i)
    for i in range(1, num1 + 1):
        q1.append(3 * i - 2)
    for i in range(1, num2 + 1):
        q0.append(3 * i - 1)
elif node0 <= num0:
    for i in range(1, node0 + 1):
        q0.append(3 * i)
    for i in range(node0 + 1, num0 + 1):
        q.append(3 * i)
    for i in range(1, num1 + 1):
        q1.append(3 * i - 2)
    for i in range(1, num2 + 1):
        q.append(3 * i - 1)
else:
    for i in range(1, node1 + 1):
        q1.append(3 * i)
    for i in range(node1 + 1, num0 + 1):
        q.append(3 * i)
    for i in range(1, num1 + 1):
        q.append(3 * i - 2)
    for i in range(1, num2 + 1):
        q0.append(3 * i - 1)

for i in range(N):
    if color[i] == 0:
        Q = q0 if q0 else q
        ans[i] = Q.pop()
    else:
        Q = q1 if q1 else q
        ans[i] = Q.pop()
print(*ans)