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.buffer.readline
from collections import deque

def main():
    N = int(input())
    edge = [[] for _ in range(N)]
    for _ in range(N-1):
        a,b = map(int,input().split())
        edge[a-1].append(b-1)
        edge[b-1].append(a-1)

    
    use = [-1 for _ in range(N)]
    d = [[0],[]]
    use[0] = 0
    q = deque([0])
    while q:
        now = q.popleft()
        for fol in edge[now]:
            if use[fol] == -1:
                use[fol] = (use[now]+1)%2
                q.append(fol)
                d[use[fol]].append(fol)

    ans = [0 for _ in range(N)]
    b = [False for _ in range(N)]
    if len(d[0]) <= N//3:
        l = len(d[0])
        for i in range(l):
            ans[d[0][i]] = 3*i+3
            b[3*i+2] = True
        t = 0
        for num in d[1]:
            while b[t]:
                t += 1
            ans[num] = t+1
            b[t] = True
    elif len(d[1]) <= N//3:
        l = len(d[1])
        for i in range(l):
            ans[d[1][i]] = 3*i+3
            b[3*i+2] = True
        t = 0
        for num in d[0]:
            while b[t]:
                t += 1
            ans[num] = t+1
            b[t] = True
    else:
        l = len(d[0])
        t = 0
        for i in range(l):
            if i < (N-1)//3+1:
                ans[d[0][i]] = 3*i+1
                b[3*i] = True
            else:
                ans[d[0][i]] = 3*t+3
                b[3*t+2] = True
                t += 1
        l = len(d[1])
        for i in range(l):
            if i < (N-2)//3+1:
                ans[d[1][i]] = 3*i+2
                b[3*i+1] = True
            else:
                ans[d[1][i]] = 3*t+3
                b[3*t+2] = True
                t += 1
    
    print(*ans)

if __name__ == "__main__":
    main()