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.*;

class Main {

	public static int[] color;
	public static int red, blue;
	public static List<Integer>[] e;

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

		e = new List[n];
		for (int i = 0; i < n; i++) e[i] = new ArrayList<>();
		for (int i = 0; i < n - 1; i++) {
			int v = sc.nextInt() - 1;
			int w = sc.nextInt() - 1;
			e[v].add(w);
			e[w].add(v);
		}

		color = new int[n];
		color[0] = 1;
		red = 1;
		blue = 0;

		dfs(0);
		StringBuilder sb = new StringBuilder();
		int a1 = 1, a2 = 2, a3 = 3;
		if (red > n / 3 && blue > n / 3) {
			for (int i = 0; i < n; i++) {
				if (color[i] == 1) {
					if (a1 <= n) {
						sb.append(a1);
						a1 += 3;
					} else {
						sb.append(a3);
						a3 += 3;
					}
				} else {
					if (a2 <= n) {
						sb.append(a2);
						a2 += 3;
					} else {
						sb.append(a3);
						a3 += 3;
					}
				}
				sb.append(" ");
			}
		} else if (red < blue) {
			for (int i = 0; i < n; i++) {
				if (color[i] == 1) {
					sb.append(a3);
					a3 += 3;
				} else {
					if (a2 <= n) {
						sb.append(a2);
						a2 += 3;
					} else if (a1 <= n) {
						sb.append(a1);
						a1 += 3;
					} else {
						sb.append(a3);
						a3 += 3;
					}
				}
			}
			sb.append(" ");
		} else {
			for (int i = 0; i < n; i++) {
				if (color[i] == 2) {
					sb.append(a3);
					a3 += 3;
				} else {
					if (a1 <= n) {
						sb.append(a1);
						a1 += 3;
					} else if (a2 <= n) {
						sb.append(a2);
						a2 += 3;
					} else {
						sb.append(a3);
						a3 += 3;
					}
				}
				sb.append(" ");
			}
		}

		System.out.println(sb.deleteCharAt(sb.length() - 1).toString());
	}

	private static void dfs(int v) {
		for (int i = 0; i < e[v].size(); i++) {
			if (color[e[v].get(i)] != 0) continue;
			if (color[v] == 1) {
				color[e[v].get(i)] = 2;
				blue++;
			} else if (color[v] == 2) {
				color[e[v].get(i)] = 1;
				red++;
			}
			dfs(e[v].get(i));
		}
	}
}