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;

int n, a, b, col = 1;
vector<int> adj[200001];
int color[200001];
int ans[200001];

void dfs(int cur) {
	for (auto nxt : adj[cur]) {
		if (color[nxt] == -1) {
			color[nxt] = !color[cur];
			dfs(nxt);
		}
	}
}

int main() {
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	memset(color, -1, sizeof(color));
	color[1] = 0;
	dfs(1);

	for (int i = 1; i <= n; i++)
		color[i] ? a++ : b++;
	if (a > b) swap(a, b), col = 0;

	if (a <= n / 3) {
		int temp1 = 3, temp2 = 1;
		for (int i = 1; i <= n && temp2 <= n; i++) {
			if (color[i] == col) 
				ans[i] = temp1, temp1 += 3;
			else if (temp2 % 3 == 1)
				ans[i] = temp2, temp2 += 1;
			else ans[i] = temp2, temp2 += 2;
		}
		for (int i = 1; i <= n; i++) {
			if (!ans[i])
				ans[i] = temp1, temp1 += 3;
		}
	} else {
		int temp1 = 1, temp2 = 2, temp3 = 3;
		for (int i = 1; i <= n && temp1 <= n; i++) {
			if (color[i] == col)
				ans[i] = temp1, temp1 += 3;
		}
		for (int i = 1; i <= n && temp2 <= n; i++) {
			if (color[i] != col)
				ans[i] = temp2, temp2 += 3;
		}
		for (int i = 1; i <= n; i++) {
			if (!ans[i]) ans[i] = temp3, temp3 += 3;
		}
	}
	for (int i = 1; i <= n; i++)
		cout << ans[i] << ' ';
}