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 P = pair<int,int>;
constexpr ll INF = 9e18;
constexpr int inf = 1e9;
constexpr double INFD = 1e100;
constexpr ll mod = 1000000007;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
// ---------------------------------------------------------------------------

int main(){
  int N;
  cin >> N;
  vector<vector<int>> G(N);
  for(int i=0; i<N-1; i++){
    int a,b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  vector<int> dist(N,-1);
  queue<int> que;
  que.push(0);
  dist[0] = 0;
  while(que.size()){
    int v = que.front();
    que.pop();
    for(int nv: G[v]){
      if(dist[nv] != -1) continue;
      dist[nv] = dist[v] + 1;
      que.push(nv);
    }
  }
  int cnt_0=0,cnt_1=0,cnt_2=0;
  for(int i=0; i<N; i++){
    if(dist[i]%3==0){
      cnt_0++;
    }else if(dist[i]%3==1){
      cnt_1++;
    }else{
      cnt_2++;
    }
  }
  if(cnt_2 == 0){
    for(int i=0; i<N; i++){
      cout << i+1 << (i!=N-1 ? " " : "\n");
    }
    return 0;
  }
  if(max({cnt_0,cnt_1,cnt_2})-min({cnt_0,cnt_1,cnt_2}) > 1){
    cout << -1 << endl;
  }else{
    int id_0=3,id_1=1,id_2=2;
    if(cnt_1>cnt_2){
      for(int i=0; i<N; i++){
        if(dist[i]%3 == 0){
          cout << id_0 << " ";
          id_0 += 3;
        }else if(dist[i]%3 == 1){
          cout << id_1 << " ";
          id_1 += 3;
        }else{
          cout << id_2 << " ";
          id_2 += 3;
        }
      }
    }else{
      for(int i=0; i<N; i++){
        if(dist[i]%3 == 0){
          cout << id_0 << " ";
          id_0 += 3;
        }else if(dist[i]%3 == 1){
          cout << id_2 << " ";
          id_2 += 3;
        }else{
          cout << id_1 << " ";
          id_1 += 3;
        }
      }
    }
    cout << endl;
  }
  return 0;
}