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

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

import queue

def main():
    N=I()
    adj=[[] for _ in range(N)]
    mod1=range(1,N+1,3)
    n1=0
    mod2=range(2,N+1,3)
    n2=0
    mod12=list(mod1)+list(mod2)
    n12=0
    mod3=range(3,N+1,3)
    n3=0
    
    p=[0]*N
    
    for i in range(N-1):
        a,b=MI()
        a-=1
        b-=1
        adj[a].append(b)
        adj[b].append(a)
     
    
    q=queue.Queue()
    q.put((0,-1,0))#node,par,depth
    d=[0]*N#深さをメモ
            
        
    while not q.empty():
        v,par,depth=q.get()
        d[v]=depth%2
                
        for i in range(len(adj[v])):
            nxt=adj[v][i]
            if nxt!=par:
                q.put((nxt,v,depth+1))
                
    dmod=[0,0]
    for i in range(N):
        dmod[d[i]]+=1
        
    if dmod[0]<=(N//3):
        for i in range(N):
            if d[i]==0:
                p[i]=mod3[n3]
                n3+=1
            else:
                if n12<len(mod12):
                    p[i]=mod12[n12]
                    n12+=1
                else:
                    p[i]=mod3[n3]
                    n3+=1
    elif dmod[1]<=(N//3):
        for i in range(N):
            if d[i]==1:
                p[i]=mod3[n3]
                n3+=1
            else:
                if n12<len(mod12):
                    p[i]=mod12[n12]
                    n12+=1
                else:
                    p[i]=mod3[n3]
                    n3+=1         
    
    else:
        for i in range(N):
            if d[i]==1:
                if n1<len(mod1):
                    p[i]=mod1[n1]
                    n1+=1
                else:
                    p[i]=mod3[n3]
                    n3+=1
            else:
                if n2<len(mod2):
                    p[i]=mod2[n2]
                    n2+=1
                else:
                    p[i]=mod3[n3]
                    n3+=1     
    
                
    print(' '.join(map(str, p)))
        
        
    
    
        
                
            
        
        
    

main()