Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")
#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
#include <bitset>
#include <cmath>
#include <limits>
#include <iostream>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <iomanip>
#include <functional>
#include <complex>

using namespace std;

using ll=long long int;
using pii=pair<int,int>;
using pil=pair<int,ll>;
using pli=pair<ll,int>;
using pll=pair<ll,ll>;
using psi=pair<string,int>;
using pis=pair<int,string>;
using psl=pair<string,ll>;
using pls=pair<ll,string>;
using pss=pair<string,string>;



template<typename T> using vc=vector<T>;
template<typename T> using vvc=vector<vector<T>>;
template<typename T> using vvvc=vector<vector<vector<T>>>;
template<typename T> using vvvvc=vector<vvvc<T>>;
template<typename T> using vvvvvc=vector<vvvvc<T>>;

template<class T,class U>inline constexpr bool chmin(T&a,const U b){if(a<=b)return false;a=b;return true;}
template<class T,class U>inline constexpr bool chmax(T&a,const U b){if(a>=b)return false;a=b;return true;}
#define bit(n,k) ( (n>>k)&1 )

inline void bin101(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr); 
    cout<<fixed<<setprecision(20);
}

//デバッグ
template<typename T>
inline void vout(const vector<T> &v){
    cout<<"\nstart\n";
    const int sz=v.size();
    for(int i=0;i<sz;i++){
        cout<<i<<" "<<v[i]<<'\n';
    }
    cout<<"end\n"<<endl;
}
//デバッグ
template<typename T>
inline void vout(const vvc<T> &v){
    cout<<"\nstart\n";
    const int sz=v.size();
    for(int i=0;i<sz;i++){
        int ssz=v[i].size();
        for(int j=0;j<ssz;j++){
            cout<<i<<" "<<j<<" "<<v[i][j]<<'\n';
        }
    }
    cout<<"end\n"<<endl;
}
//デバッグ
template<typename T>
inline void vout(const vvvc<T> &v){
    cout<<"\nstart\n";
    const int sz=v.size();
    for(int i=0;i<sz;i++){
        int ssz=v[i].size();
        for(int j=0;j<ssz;j++){
            int sssz=v[i][j].size();
            for(int k=0;k<sssz;k++){
                cout<<i<<" "<<j<<" "<<k<<" "<<v[i][j][k]<<'\n';
            }
        }
    }
    cout<<"end\n"<<endl;
}


//pair cout
template<typename T, typename U>
inline ostream &operator<<(ostream &os,const pair<T,U> &p) {
  os<<p.first<<" "<<p.second;
  return os;
}

//pair cin
template<typename T, typename U>
inline istream &operator>>(istream &is,pair<T,U> &p) {
  is>>p.first>>p.second;
  return is;
}

//ソート
template<typename T>
inline void vsort(vector<T> &v){
    sort(v.begin(),v.end());
}

//逆順ソート
template<typename T>
inline void rvsort(vector<T> &v){
  sort(v.rbegin(),v.rend());
}

//要素数a 初期値x
template<typename T>
inline vector<T> vmake(int a,T x){
  return vector<T>(a,x);
}

//data[a][b] 初期値x
template<typename T>
inline vector<vector<T>> vmake(int a,int b,T x){
  return vector<vector<T>>(a, vector<T>(b,x));
}

//data[a][b][c] 初期値x
template<typename T>
inline vector<vector<vector<T>>> vmake(int a,int b,int c,T x){
  return vector<vector<vector<T>>> (a, vector<vector<T>>(b, vector<T>(c, x)));
}

//data[a][b][c][d]　初期値x
template<typename T>
inline vector<vector<vector<vector<T>>>> vmake(int a,int b,int c,int d,T x){
  return vector<vector<vector<vector<T>>>> (a,vvvc<T>(b,vvc<T>(c,vc<T>(d,x))));
}

//data[a][b][c][d][e] 初期値x
template<typename T>
inline vvvvvc<T> vmake(int a,int b,int c,int d,int e,T x){
  return vvvvvc<T> (a,vvvvc<T>(b,vvvc<T>(c,vvc<T>(d,vc<T>(e,x)))));
}

//1ビットの数を返す
inline int popcount(int x){
  return __builtin_popcount(x);
}
//1ビットの数を返す
inline int popcount(ll x){
  return __builtin_popcountll(x);
}

//queのfront() pop()
template<typename T>
inline T pop(queue<T> &que){
    T x=que.front();
    que.pop();
    return x;
}

//priority_que top() pop()
template<typename T>
inline T pop(priority_queue<T> &que){
  T x=que.top();
  que.pop();
  return x;
}

//stack top() pop()
template<typename T>
inline T pop(stack<T> &st){
  T x=st.top();
  st.pop();
  return x;
}

#define intsz(x) ( (int)x.size()  )

/*
満たすものの個数を返す
mode:0 x未満
mode:1 x以下の数
mode:2 x以上の数
mode:3 x超
mode:4 x
*/
template<typename T>
inline int count_bound(vector<T> &v,T x,int mode){
    switch(mode){
        case 0:
            return lower_bound(v.begin(),v.end(),x)-v.begin();
        case 1:
            return upper_bound(v.begin(),v.end(),x)-v.begin();
        case 2:
            return v.end()-lower_bound(v.begin(),v.end(),x);
        case 3:
            return v.end()-upper_bound(v.begin(),v.end(),x);
        case 4:
            return upper_bound(v.begin(),v.end(),x)-lower_bound(v.begin(),v.end(),x);
    }
}

/*
mode:0 xより小さい数で最大の数
mode:1 x以下の数で最大の数
mode:2 x以上の数で最小の数
mode:3 xより大きい数で最小の数
*/
template<typename T>
inline T value_bound(vector<T> &v,T x,int mode){
    switch(mode){
        case 0:
            return *(--lower_bound(v.begin(),v.end(),x));
        case 1:
            return *(--upper_bound(v.begin(),v.end(),x));
        case 2:
            return *lower_bound(v.begin(),v.end(),x);
        case 3:
            return *upper_bound(v.begin(),v.end(),x);
    }
}



constexpr int MAX=1<<30;
constexpr ll INF=1LL<<62;
constexpr ll MOD=1e9+7;
int dx[]={1,-1,0,0},dy[]={0,0,-1,1};

//+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
//+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
//+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


signed main(){
    bin101();

    cout<<-1<<endl;
}