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;
using ll = long long;

vector<vector<int>> g;
vector<int> dep;

void dfs(int v, int pr) {
	for (auto u : g[v]) {
		if (u != pr) {
			dep[u] = dep[v] + 1;
			dfs(u, v);
		}
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int n;
	cin >> n;

	g.resize(n);

	for (int i = 1; i < n; i++) {
		int v, u;
		cin >> v >> u;
		v--, u--;
		g[v].push_back(u);
		g[u].push_back(v);
	}

	dep.resize(n);

	dfs(0, 0);

	array<vector<int>, 2> vers;
	for (int i = 0; i < n; i++) {
		vers[dep[i] % 2].push_back(i);
	}
	if (vers[1].size() < vers[0].size()) {
		swap(vers[0], vers[1]);
	}

	array<vector<int>, 3> free;
	for (int i = 1; i <= n; i++) {
		free[i % 3].push_back(i);
	}

	vector<int> ans(n, -1);

	if (vers[0].size() <= free[0].size()) {
		for (auto el : vers[0]) {
			ans[el] = free[0].back();
			free[0].pop_back();
		}
	} else {
		while (free[1].size()) {
			ans[vers[0].back()] = free[1].back();
			free[1].pop_back();
			vers[0].pop_back();
		}

		while (free[2].size()) {
			ans[vers[1].back()] = free[2].back();
			free[2].pop_back();
			vers[1].pop_back();
		}
	}

	for (int i = 0; i < n; i++) {
		if (ans[i] == -1) {
			int j = 0;
			while (!free[j].size()) {
				j++;
			}
			ans[i] = free[j].back();
			free[j].pop_back();
		}
	}

	for (auto el : ans) {
		cout << el << ' ';
	}
}