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 CPP):

Solution:
#include<bits/stdc++.h>
#define endl enjoy_codeforces
int main(){
    int n;std::cin>>n;
    std::vector<int>sz={n/3,(n+2)/3,(n+1)/3};
    std::vector<std::vector<int>>g(n);
    for(int i=0;i<n-1;i++){
        int u,v;std::cin>>u>>v;u--,v--;
        g.at(u).push_back(v);
        g.at(v).push_back(u);
    }
    std::vector<int>a(n,-1);
    std::queue<int>que;
    que.push(0);
    a.at(0)=0;
    while(!que.empty()){
        int x=que.front();que.pop();
        for(int y:g.at(x)){
            if(a.at(y)==-1){
                a.at(y)=!a.at(x);
                que.push(y);
            }
        }
    }
    std::vector<int>b,c;
    for(int i=0;i<n;i++)(a.at(i)?b:c).push_back(i);
    std::vector<int>p(n);
    for(int x:{1,2}){
        if(b.size()<c.size())std::swap(b,c);
        for(int i=x;i<=n;i+=3){
            p.at(b.back())=i;
            b.pop_back();
        }
    }
    for(int i=0,j=3;i<n;i++){
        if(p.at(i)==0){
            p.at(i)=j;
            j+=3;
        }
    }
    for(int i=0;i<n;i++){
        std::cout<<(i?" ":"")<<p.at(i);
    }
    std::cout<<'\n';
}
/*
二部グラフです。
余りが 1 同士、2 同士が対岸にこないようにします。
大きい方に 1 を入れて、残りの内大きい方に 2 を入れれば良いです。
*/