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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;
import java.util.ArrayDeque;
import java.util.ArrayList;
public class Main {
  public static void main (String args[]) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int three = N/3;
    ArrayList<Integer>[] edge  = new ArrayList[N];
    for(int i = 0; i < N; i++){
      edge[i] = new ArrayList<>();
    }
    while(sc.hasNext()){
      int from = sc.nextInt()-1;
      int to = sc.nextInt()-1;
      edge[from].add(to);
      edge[to].add(from);
    }
    int ans[] = new int[N];
    int depth[] = new int[N];
    depth[0] = 1;
    int ki, gu;
    ki = gu = 0;
    ArrayDeque<Integer> adq = new ArrayDeque<>();
    adq.add(0);
    while(adq.size()>0){
      int now = adq.pollFirst();
      if(depth[now]==1){
        gu++;
      } else {
        ki++;
      }
      for(int next : edge[now]){
        if(depth[next]==0){
          depth[next] = -depth[now];
          adq.addLast(next);
        }
      }
    }
    boolean check[] = new boolean[N];
    int k, l, m;
    k = l = 0;
    m = 1;
    if(ki<=three){
      for(int i=0;i<N;i++){
        if(depth[i]==-1) {
            ans[i]=3*m;
            check[ans[i]-1]=true;
            m++;
        }
      }
    }
    else if(gu<=three){
      for(int i=0;i<N;i++){
        if(depth[i]==1) {
            ans[i]=3*m;
            check[ans[i]-1]=true;
            m++;
        }
      }
    }
    else {
      for(int i=0;i<N;i++){
        if(depth[i]==-1&&k<=three) {
          ans[i]=3*k+1;
          check[ans[i]-1]=true;
          k++;
        }
        else if(depth[i]==1&&l<=three){
          ans[i]=3*l+2;
          check[ans[i]-1]=true;
          l++;
        }
      }
    }
    int c = 0;
    for(int i=0;i<N;i++){
      if(ans[i]!=0){
        System.out.print(ans[i]+" ");
      } else {
        while(check[c]){ 
          c++;
        }
        System.out.print((c+1)+" ");
        c++;
      }
    }
    for(int i = 0; i < N; i++){
      System.err.println(depth[i]);
    }
  }
}