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 <bits/stdc++.h> 

using namespace std;
using ll=long long;
using ull=unsigned long long;
using pii=pair<int,int>;

#define INF LONG_MAX
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.end(),a.begin()
#define rep(i,N) for(int i=0;i<N;i++)

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N;
  cin>>N;
  vector<int>a(N-1),b(N-1);
  for(int i=0;i<N-1;i++)cin>>a[i]>>b[i];
  vector<vector<int>>v(N);
  for(int i=0;i<N-1;i++){
    a[i]--;
    b[i]--;
    v[a[i]].push_back(b[i]);
  }
  vector<int>p(N,0);
  int one=N/3;
  int two=N/3;
  if(N%3>=1)one++;
  if(N%3>=2)two++;
  [&](auto f){return [=](auto...args){return f(f,args...);};}([&](auto f,int now,int par,int c)->int{
      if(c==1){
        if(one==0){
          p[now]=3;
        }
        else{
          p[now]=1;
          one--;
        }
      }
      else{
        if(two==0){
          p[now]=3;
        }
        else{
          p[now]=2;
          two--;
        }
      }
      if(c==1)c=2;
      else if(c==2)c=1;
      for(auto e:v[now]){
        if(par==e)continue;
        f(f,e,now,c);
      }
  })(0,-1,1);
  int A=0,B=0,C=0;
  for(auto it:p){
    if(it==1){
      cout<<3*A+it<<" ";
      A++;
    }
    if(it==2){
      cout<<3*B+it<<" ";
      B++;
    }
    if(it==3){
      cout<<3*C+it<<" ";
      C++;
    }
  }
  return 0;
}