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
input=sys.stdin.readline
sys.setrecursionlimit(100000000)

def main():
    n=int(input())
    tree=[[] for i in range(n)]
    for _ in range(n-1):
       a,b=map(int,input().split())
       a-=1
       b-=1
       tree[a].append(b)
       tree[b].append(a)
    
    depth=[-1]*n
    depth[0]=0

    def dfs(i):
        for e in tree[i]:
            if depth[e]==-1:
                depth[e]=(depth[i]+1)%2
                dfs(e)
    dfs(0)
    o=sum(depth)
    e=n-o

    if n%3==0:
        i = n//3
        j = i
        k = i
    elif n%3==1:
        i = n//3
        j = i+1
        k = i
    else:
        i = n//3
        j = i+1
        k = j

    ans=[0]*n
    if 3*e > n and 3*o > n:
        for l in range(n):
            if depth[l]==0:
                if j==0:
                    ans[l]=3*i
                    i-=1
                else:
                    ans[l]=3*j-2
                    j-=1
            else:
                if k==0:
                    ans[l]=3*i
                    i-=1
                else:
                    ans[l]=3*k-1
                    k-=1
    
    elif 3*e <= n:
        for l in range(n):
            if depth[l]==0:
                ans[l]=3*i
                i-=1
            else:
                if j==0 and k==0:
                    ans[l]=3*i
                    i-=1
                elif j==0:
                    ans[l]=3*k-1
                    k-=1
                else:
                    ans[l]=3*j-2
                    j-=1
    
    else:
        for l in range(n):
            if depth[l]==1:
                ans[l]=3*i
                i-=1
            else:
                if j==0 and k==0:
                    ans[l]=3*i
                    i-=1
                elif j==0:
                    ans[l]=3*k-1
                    k-=1
                else:
                    ans[l]=3*j-2
                    j-=1
    print(*ans)
    
if __name__=='__main__':
    main()