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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

sys.setrecursionlimit(10**9)


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

flag = [0]*n
def dfs(cur, par,f):
    flag[cur] = f
    for to in G[cur]:
        if to != par:
            dfs(to, cur, f^1)

dfs(0, -1, 0)

a = sum(flag)
b = n-a

one = []
two = []
three = []
for i in range(1, n+1):
    if i%3 == 0:
        three.append(i)
    elif i%3 == 1:
        one.append(i)
    else:
        two.append(i)
k = n//3
ans = [0]*n
if k<a and k<b:
    for i in range(n):
        if flag[i]:
            if one:
                ans[i] = one.pop()
            else:
                ans[i] = three.pop()
        else:
            if two:
                ans[i] = two.pop()
            else:
                ans[i] = three.pop()
else:
    #1
    if k >= a:
        for i in range(n):
            if flag[i]:
                ans[i] = three.pop()
        res = one + two + three
        for i in range(n):
            if not flag[i]:
                ans[i] = res.pop()
    # 0
    else:
        for i in range(n):
            if not flag[i]:
                ans[i] = three.pop()
        res = one + two + three
        for i in range(n):
            if flag[i]:
                ans[i] = res.pop()
print(*ans)