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
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
def resolve():
    N = ir()
    edges = [[] for _ in range(N)]
    for i in range(N-1):
        a, b = lr()
        edges[a-1].append(b-1)
        edges[b-1].append(a-1)
    one = N//3+min(1, N%3)
    two = N//3+N%3//2
    thr = N//3
    ans = [0]*N
    ball = [0]*N
    cnt = 0
    dist = [-1]*N
    s = [0]
    dist[0] = 0
    while len(s) > 0:
        v = s.pop()
        for j in edges[v]:
            if dist[j] == -1:
                dist[j] = dist[v]+1
                if dist[j]%2 == 1:
                    ball[j] = 1
                    cnt += 1
                s.append(j)
    if cnt > N/3 and N-cnt > N/3:
        for i in range(N):
            if ball[i] == 1:
                ans[i] = 3*one-2
                one -= 1
            else:
                ans[i] = 3*two-1
                two -= 1
        for i in range(N):
            if ans[i] <= 0:
                ans[i] = 3*thr
                thr -= 1
    elif cnt <= N/3:
        for i in range(N):
            if ball[i] == 1:
                ans[i] = 3*thr
                thr -= 1
        for i in range(N):
            if ans[i] <= 0:
                if thr > 0:
                    ans[i] = 3*thr
                    thr -= 1
                elif two > 0:
                    ans[i] = 3*two-1
                    two -= 1
                else:
                    ans[i] = 3*one-2
                    one -= 1
    else:
        for i in range(N):
            if ball[i] == 0:
                ans[i] = 3*thr
                thr -= 1
        for i in range(N):
            if ans[i] <= 0:
                if thr > 0:
                    ans[i] = 3*thr
                    thr -= 1
                elif two > 0:
                    ans[i] = 3*two-1
                    two -= 1
                else:
                    ans[i] = 3*one-2
                    one -= 1
    print(*ans, sep=' ') 
resolve()