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:
n = int(input())
con = [set() for _ in range(n)]
tree = [0]*n

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

checked = set([1])
stack = list(con[0])
stack_ = []
while len(checked) < n:
    while stack:
        x = stack.pop()
        if x in checked:
            continue
        checked.add(x)
        stack_.extend(con[x-1])
        tree[x-1] = 1
        
    while stack_:
        x = stack_.pop()
        if x in checked:
            continue
        checked.add(x)
        stack.extend(con[x-1])

mod0 = []
mod1 = []
mod2 = []
for i in range(n):
    if (i+1)%3 == 0:
        mod0.append(i+1)
    elif (i+1)%3 == 1:
        mod1.append(i+1)
    else:
        mod2.append(i+1)

ans = [0]*n
sum_tree = sum(tree)
if sum_tree < n/3:
    for i in range(n):
        if tree[i] == 0:
            if mod1:
                ans[i] = mod1.pop()
            elif mod2:
                ans[i] = mod2.pop()
            else:
                ans[i] = mod0.pop()
        else:
            ans[i] = mod0.pop()
elif sum_tree > n*2/3:
    for i in range(n):
        if tree[i] == 1:
            if mod1:
                ans[i] = mod1.pop()
            elif mod2:
                ans[i] = mod2.pop()
            else:
                ans[i] = mod0.pop()
        else:
            ans[i] = mod0.pop() 
else:
    for i in range(n):
        if tree[i] == 0:
            if mod1:
                ans[i] = mod1.pop()
            else:
                ans[i] = mod0.pop()
        else:
            if mod2:
                ans[i] = mod2.pop()
            else:
                ans[i] = mod0.pop()
            
print(" ".join(map(lambda x:str(x), ans)))