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
sys.setrecursionlimit(10**7)
input = sys.stdin.readline
N = int(input())
AB = [tuple(map(int,input().split())) for i in range(N-1)]
if N==2:
    print(1,2)
    exit()
es = [[] for i in range(N)]
for a,b in AB:
    a,b = a-1,b-1
    es[a].append(b)
    es[b].append(a)

depth = [-1]*N
depth[0] = 0
def dfs(v,p=-1):
    for to in es[v]:
        if to==p: continue
        depth[to] = depth[v] + 1
        dfs(to,v)
dfs(0)

g1 = []
g2 = []
for i,d in enumerate(depth):
    if d%2:
        g2.append(i)
    else:
        g1.append(i)
if len(g1) < len(g2):
    g1,g2 = g2,g1

ans = [None] * N
if len(g2) <= N//3:
    for n in range(1,N+1):
        if n%3==0:
            if g2:
                i = g2.pop()
            else:
                i = g1.pop()
        else:
            i = g1.pop()
        ans[i] = n
else:
    for n in range(1,N+1,3):
        i = g1.pop()
        ans[i] = n
    for n in range(2,N+1,3):
        i = g2.pop()
        ans[i] = n
    for n in range(3,N+1,3):
        if g1:
            i = g1.pop()
        else:
            i = g2.pop()
        ans[i] = n
print(*ans)