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 os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353

N = int(sys.stdin.buffer.readline())
AB = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N - 1)]
# AB = generate_tree(N)


graph = [[] for _ in range(N)]
for a, b in AB:
    a -= 1
    b -= 1
    graph[a].append(b)
    graph[b].append(a)

bp = [-1] * N
stack = [(0, None, 0)]
while stack:
    v, p, b = stack.pop()
    bp[v] = b
    b += 1
    b &= 1
    for u in graph[v]:
        if u == p:
            continue
        stack.append((u, v, b))

cnt0 = N // 3
cnt1 = (N + 2) // 3
cnt2 = (N + 1) // 3
vs0 = [i for i, b in enumerate(bp) if b == 0]
vs1 = [i for i, b in enumerate(bp) if b == 1]
ans_mod3 = [-1] * N
if bp.count(0) <= N // 3:
    for i in range(len(vs0)):
        ans_mod3[vs0.pop()] = 0
        cnt0 -= 1
elif bp.count(1) <= N // 3:
    for i in range(len(vs1)):
        ans_mod3[vs1.pop()] = 0
        cnt0 -= 1
else:
    for i in range(cnt1):
        ans_mod3[vs0.pop()] = 1
        cnt1 -= 1
    for i in range(cnt2):
        ans_mod3[vs1.pop()] = 2
        cnt2 -= 1

for _ in range(cnt0):
    if vs0:
        ans_mod3[vs0.pop()] = 0
    elif vs1:
        ans_mod3[vs1.pop()] = 0
for _ in range(cnt1):
    if vs0:
        ans_mod3[vs0.pop()] = 1
    elif vs1:
        ans_mod3[vs1.pop()] = 1
for _ in range(cnt2):
    if vs0:
        ans_mod3[vs0.pop()] = 2
    elif vs1:
        ans_mod3[vs1.pop()] = 2

st = [3, 1, 2]
ans = []
for a in ans_mod3:
    ans.append(st[a])
    st[a] += 3
print(*ans)