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:
from collections import deque


n = int(input())

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


root = 0
visited = [-1] * n
def dfs(root):
    q = deque([root])
    visited[root] = 0
    while q:
        v = q.pop()
        for nxt_v in tree[v]:
            if visited[nxt_v] != -1:
                continue
            else:
                visited[nxt_v] = (visited[v] + 1) % 2
                q.append(nxt_v)
dfs(0)
cnt0 = 0
cnt1 = 0
for i in range(n):
    if visited[i] == 0:
        cnt0 += 1
    else:
        cnt1 += 1

num1 = n // 3 
num2 = n // 3
num3 = n // 3
if n % 3 >= 1:
   num1 += 1
if n % 3 >= 2:
   num2 += 1
    
ans = [0] * n

set1 = set()
set2 = set()
set3 = set()
for i in range(1, n + 1):
    if i % 3 == 0:
        set3.add(i)
    elif i % 3 == 1:
        set1.add(i)
    elif i % 3 == 2:
        set2.add(i)
        
if cnt0 <= num3:
    for i in range(n):
        if visited[i] == 0:
            ans[i] = set3.pop()
        else:
            if set1:
                ans[i] = set1.pop() 
            elif set2:
                ans[i] = set2.pop()         
            elif set3:
                ans[i] = set3.pop()  

elif cnt1 <= num3:
    for i in range(n):
        if visited[i] == 1:
            ans[i] = set3.pop()
        else:
            if set1:
                ans[i] = set1.pop() 
            elif set2:
                ans[i] = set2.pop()         
            elif set3:
                ans[i] = set3.pop()  
else:
    num147 = 0
    num258 = 0
    num369 = 0
    for i in range(n):
        if visited[i] == 0:
            num = num147 * 3 + 1
            if num > n:
                num = num369 * 3 + 3
                ans[i] = num
                if num > n:
                    error
                    exit()
                set3.remove(num)
                num369 += 1
                continue
            ans[i] = num # 1, 4, 7, 10, ...
            set1.remove(num)
            num147 += 1
        else:
            num = num258 * 3 + 2 
            if num > n:
                num = num369 * 3 + 3
                if num > n:
                    error
                    exit()
                ans[i] = num
                set3.remove(num)
                num369 += 1
                continue
            ans[i] = num # 2, 5, 8, 11, ...
            set2.remove(num)
            num258 += 1
     
    for i in range(n):
        if ans[i] == 0:
            if set1:
                ans[i] = set1.pop()
            elif set2:
                ans[i] = set2.pop()
            elif set3:
                ans[i] = set3.pop()
print(*ans)