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)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<queue>
#include<cmath>
#include<cstdio>
#include<tuple>
#include<bitset>

using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
#define ALL(x) x.begin(),x.end()
#define ll long long
const int inf=1000000007;
const int MOD=1000000007;
struct Tree{
  bool tag;
  bool comp;
  vector<int> edge;
  int count(){
    int num=edge.size();
    return num;
  }
};
signed main(){
  int n;cin>>n;
  vector<Tree> tree(n+1);
  rep(i,n-1){
    int a,b;cin>>a>>b;
    tree[a].edge.push_back(b);
    tree[b].edge.push_back(a);
  }
  rep(i,n+1){
    //tree[i].tag=true;
    tree[i].comp=true;
    //cout<<i<<endl;
  }
  //return 0;
  //tree[0].comp=true;
  queue<int> que;que.push(1);
  tree[1].comp=false;
  tree[1].tag=true;
  while(!que.empty()){
    int buf=que.front();que.pop();
    int num=tree[buf].count();
    rep(i,num){
      if(tree[tree[buf].edge[i]].comp){
        tree[tree[buf].edge[i]].comp=false;
        que.push(tree[buf].edge[i]);
        tree[tree[buf].edge[i]].tag=!tree[buf].tag;
      }
    }
  }
  int T=0;int F=0;
  rep1(i,n){
    if(tree[i].tag)T++;
    else F++;
  }
  int buf=max(T,F);
  F=min(T,F);
  T=buf;
  int one=0;int two=0;int three=0;
  if(T<=n/3){
    rep1(i,n){
      if(tree[i].tag){
        three++;
        cout<<three*3;
      }
      else{
        one++;
        if(one%3==0)one++;
        if(one>n){
          three++;
          cout<<three*3;
        }
        else cout<<one;
      }
      cout<<" ";
    }
  }
  else if(F<=n/3){
    rep1(i,n){
      if(!tree[i].tag){
        three++;
        cout<<three*3;
      }
      else{
        one++;
        if(one%3==0)one++;
        if(one>n){
          three++;
          cout<<three*3;
        }
        else cout<<one;
      }
      cout<<" ";
    }
  }
  else{
    rep1(i,n){
      if(tree[i].tag){
        if(one*3+1<=n){
          cout<<one*3+1;
          one++;
        }
        else{
          three++;
          cout<<three*3;
        }
      }
      else{
        if(two*3+2<=n){
          cout<<two*3+2;
          two++;
        }
        else{
          three++;
          cout<<three*3;
        }
      }
      cout<<" ";
    }
  }
  cout<<endl;
  return 0;
}