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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

vector<int> v[200000];
int n;

int visited[200000] = {};
int count_red(int i, int color) {
	if (visited[i])
		return 0;
	visited[i] = color;
	int r = color == 2 ? 1 : 0;
	for(int a: v[i]) {
		r += count_red(a, 3 - color);
	}
	return r;
}


void print_color(int color) {
	int pm = 3, pr = 1;
	int exhausted = 0;
	for(int i = 0; i < n; i++) {
		if (i) cout << " ";
		if (!exhausted && visited[i] == color) {
			cout << pr;
			pr += pr % 3 == 1 ? 1 : 2;
			if (pr > n)
				exhausted = 1;
		} else {
			cout << pm;
			pm += 3;
		}
	}
}

int main() {
	cin >> n;
	for(int i = 0; i < n - 1; i++) {
		int a, b;
		cin >> a >> b;
		v[a - 1].push_back(b - 1);
		v[b - 1].push_back(a - 1);
	}
	int rc = count_red(0, 1);
	int bc = n - rc;
	if (rc >= n / 3 * 2 + n % 3) {
		print_color(2);
	} else if (bc >= n / 3 * 2 + n % 3) {
		print_color(1);
	} else {
		int exhausted1 = 0, exhausted2 = 0;
		int pm = 3, pr1 = 1, pr2 = 2;
		for(int i = 0; i < n; i++) {
			if (i) cout << ' ';
			if (!exhausted1 && visited[i] == 1) {
				cout << pr1;
				pr1 += 3;
				if (pr1 > n)
					exhausted1 = 1;
			} else if (!exhausted2 && visited[i] == 2) {
				cout << pr2;
				pr2 += 3;
				if (pr2 > n)
					exhausted2 = 1;
			} else {
				cout << pm;
				pm += 3;
			}
		}
	}
	cout << endl;
}