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

rb = [-1]*n
rb[0] = 1
count1 = 1
count2 = 0
rb_m = [[0]]
for i in rb_m:
    n_rb = []
    for j in i:
        x = rb[j]
        for k in ab[j]:
            if(rb[k] != -1):
                continue
            if(x == 1):
                rb[k] = 2
                count2 += 1
            elif(x == 2):
                rb[k] = 1
                count1 += 1
            n_rb.append(k)
    set(n_rb)
    if(n_rb == []):
        break
    else:
        rb_m.append(n_rb)

ans = [0 for i in range(n)]
if(count1 > n//3 and count2 > n//3):
    ne1 = 1
    ne2 = 2
    ne3 = 3
    for i in range(n):
        if(rb[i] == 1 and ne1 <= n):
            ans[i] = ne1
            ne1 += 3
        elif(rb[i] == 2 and ne2 <= n):
            ans[i] = ne2
            ne2 += 3
        else:
            ans[i] = ne3
            ne3 += 3
elif(count2 >= 2*n//3):
    ne1 = 1
    ne2 = 2
    ne3 = 3
    for i in range(n):
        if(rb[i] == 1):
            ans[i] = ne3
            ne3 += 3
        elif(ne1 <= n):
            ans[i] = ne1
            ne1 += 3
        elif(ne2 <= n):
            ans[i] = ne2
            ne2 += 3
        else:
            ans[i] = ne3
            ne3 += 3
elif(count1 >= 2*n//3):
    ne1 = 1
    ne2 = 2
    ne3 = 3
    for i in range(n):
        if(rb[i] == 2):
            ans[i] = ne3
            ne3 += 3
        elif(ne1 <= n):
            ans[i] = ne1
            ne1 += 3
        elif(ne2 <= n):
            ans[i] = ne2
            ne2 += 3
        else:
            ans[i] = ne3
            ne3 += 3
print(*ans)