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: Wrong Answer)

import java.io.*;
import java.util.*;

public class Main {
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
//    long mod = 1000000000+7;
    int cnt2 = 0;
    ArrayList<Integer>[] tree;
    public static void main(String[] args) throws IOException {
        Main main = new Main();
//        main.solve();
        out.println(main.solve());
        out.flush();
    }
    String solve() throws IOException {
        int n = sc.nextInt();
        tree = new ArrayList[n+1];
        for(int i=0;i<=n;i++) tree[i] = new ArrayList<>();
        for(int i=0;i<n-1;i++){
            int a = sc.nextInt(), b = sc.nextInt();
            tree[a].add(b); tree[b].add(a);
        }
        int[] color = new int[n+1];
        dfs(1,1,1,color);
        Deque<Integer> one = new LinkedList<>(), two = new LinkedList<>(), three = new LinkedList<>();
        for(int i=1;i<=n;i++){
            if(i%3==1) one.add(i);
            else if(i%3==2) two.add(i);
            else three.add(i);
        }
        StringBuilder sb = new StringBuilder();
        for(int i=1;i<=n;i++){
            int ans;
            if(color[i]==1&&one.size()>0){
                ans = one.pollFirst();
            } else if(color[i]==2&&two.size()>0){
                ans = two.pollFirst();
            } else {
                if(three.size()>0) ans = three.pollFirst();
                else return "-1";
            }
            sb.append(ans).append(' ');
        }
        return sb.toString();
    }
    void dfs(int cur, int from, int cur_color, int[] color){
        color[cur] = cur_color;
        for(int to:tree[cur]){
            if(to==from) continue;
            dfs(to,cur,3-cur_color,color);
        }
    }
}