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> from[200010];
int ans[200010];
int cnt[10];

void dfs(int x, int last) {
	cnt[ans[x]]++;
	for (int i = 0; i < from[x].size(); i++) {
		int v = from[x][i];
		if (v == last) continue;
		ans[v] = 3 - ans[x], dfs(v, x);
	}
}

int main () {
	int n; scanf("%d", &n);
	int a = n / 3, b = (n + 2) / 3, c = (n + 1) / 3;
	for (int i = 1; i < n; i++) {
		int x, y; scanf("%d%d", &x, &y);
		from[x].push_back(y);
		from[y].push_back(x);
	}
	ans[1] = 1, dfs(1, 1);
	if (cnt[1] < cnt[2]) {
		for (int i = 1; i <= n; i++) {
			ans[i] = 3 - ans[i];
		}
		swap(cnt[1], cnt[2]);
	}
	if (cnt[1] <= b + a) {
		int t1 = cnt[1] - b, t2 = cnt[2] - c;
		for (int i = 1; i <= n; i++) {
			if (ans[i] == 1) {
				if (t1) t1--, ans[i] = 0;
			} else {
				if (t2) t2--, ans[i] = 0;
			}
		}
		int now[3] = {3, 1, 2};
		for (int i = 1; i <= n; i++) {
			printf("%d ", now[ans[i]]);
			now[ans[i]] += 3;
		}
		printf("\n");
		return 0;
	}
	int T = c, T1 = a - cnt[2];
	for (int i = 1; i <= n; i++) {
		if (ans[i] == 2) {
			ans[i] = 0;
		} else if (ans[i] == 1) {
			if (T) T--, ans[i] = 2;
			else if (T1) T1--, ans[i] = 0;
		}
	}
	int now[3] = {3, 1, 2};
	for (int i = 1; i <= n; i++) {
		printf("%d ", now[ans[i]]);
		now[ans[i]] += 3;
	}
	printf("\n");
	return 0;
}