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.util.*;

public class Main {

    static Scanner sc = new Scanner(System.in);
    static int n = sc.nextInt();
    static List<Integer>[] edge = new ArrayList[n];
    static boolean[] visit = new boolean[n];
//    static int[] depth

    public static void main(String[] args){
//        Scanner sc = new Scanner(System.in);

//        int n = sc.nextInt();

        for (int i=0; i<n; i++) {
            edge[i] = new ArrayList<>();
        }




        for (int i=0; i<n-1; i++) {
            int a = sc.nextInt()-1;
            int b = sc.nextInt()-1;
            edge[a].add(b);
            edge[b].add(a);
        }

        int[] ans = new int[n];

        int cnt = 1;

        for (int i=0; i<n; i+=2) {
            ans[i] = cnt++;
        }

        for (int i=1; i<n; i+=2) {
            ans[i] = cnt++;
        }

//        int cnt = 1;
//        for (int i=0; i<n; i++) {
//            for (int v: edge[i]) {
//
//            }
//        }

        for (int i=0; i<n; i++) {
            System.out.print(ans[i]+" ");
        }
//        System.out.print(edge[0]);


    }

    public static void dfs(int v) {

    }

    public class Node {
        int parent;
//        int
    }

}