Problem: 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.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		AdjacencyList[] tree = new AdjacencyList[n]; //adjacency list

		for (int i = 0; i < n; i++) {
			tree[i] = new AdjacencyList();
		}

		for (int i = 0, a, b; i < n - 1; i++) {
			a = scanner.nextInt() - 1;
			b = scanner.nextInt() - 1;
			tree[a].add(b);
			tree[b].add(a);
		}
		int[] distance = new int[n];
		for (int i = 1; i < n; i++) {
			distance[i] = -1;
		}
		dfs(tree, 0, distance);
		int even = 0;
		for (int i = 0; i < n; i++) {
			if (distance[i] % 2 == 0)
				even++;
		}

		if (even <= n / 3) {
			for (int i = 0; i < n; i++) {
				if (distance[i] % 2 == 0) {
					System.out.print(poll(0, 0, 0, n));
				} else {
					System.out.print(poll(1, 2, 0, n));
				}
				System.out.print(i == n - 1 ? "\n" : " ");
			}
		} else if (n - even <= n / 3) {
			for (int i = 0; i < n; i++) {
				if (distance[i] % 2 != 0) {
					System.out.print(poll(0, -1, -1, n));
				} else {
					System.out.print(poll(1, 2, 0, n));
				}
				System.out.print(i == n - 1 ? "\n" : " ");
			}
		} else {
			for (int i = 0; i < n; i++) {
				if (distance[i] % 2 == 0) {
					System.out.print(poll(1, 0, -1, n));
				} else {
					System.out.print(poll(2, 0, -1, n));
				}
				System.out.print(i == n - 1 ? "\n" : " ");
			}
		}

	}

	static void dfs(AdjacencyList[] tree, int v, int[] distance) {
		for (int nv : tree[v]) {
			if (distance[nv] < 0) {
				distance[nv] = distance[v] + 1;
				dfs(tree, nv, distance);
			}
		}
	}

	static int[] number = { 3, 1, 2 };

	static int poll(int first, int second, int third, int n) {
		if (number[first] <= n) {
			number[first] += 3;
			return number[first] - 3;
		}
		return poll(second, third, first, n);
	}

}

class AdjacencyList extends ArrayList<Integer> {
}