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:
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
from collections import deque

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

color = [-1] * n
def dfs(p, v, c):
    for nv in edge[v]:
        color[nv] = not c
        if nv == p:
            continue
        dfs(v, nv, not c)

dfs(-1, 0, 0)
white = sum(color)
black = n - white
if white > black:
    white, black = black, white
    color = [not item for item in color]

# Fill 0 mod 3 or not
if white <= n // 3:
    threes = [item for item in range(1, n+1) if item % 3 == 0]
    others = [item for item in range(1, n+1) if item % 3 != 0]
    array = deque(others + threes)
# Fill 1 mod 3 or 2 mod 3
else:
    threes = [item for item in range(1, n+1) if item % 3 == 0]
    twos = [item for item in range(1, n+1) if item % 3 == 2]
    ones = [item for item in range(1, n+1) if item % 3 == 1]
    array = deque(ones + threes + twos)

ans = [-1] * n
for i, is_white in enumerate(color):
    if is_white:
        ans[i] = array.pop()
    else:
        ans[i] = array.popleft()

print(*ans)