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;
vector<vector<int>> G;
vector<int> col;
void f(int u, int x) {
	col[u] = x;
	for (int v : G[u]) if (col[v] < 0) f(v, x^1);
}
int main() {
	cin >> N;
	G.resize(N);
	for (int i = 0; i < N-1; ++ i) {
		int u, v;
		cin >> u >> v;
		-- u; -- v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	col.resize(N, -1);
	f(0, 0);
	vector<vector<int>> a(3);
	vector<vector<int>> b(2);
	for (int i = 1; i <= N; ++ i) a[i%3].push_back(i);
	for (int i = 0; i < N; ++ i) b[col[i]].push_back(i);
	vector<int> r(N);
	if (b[0].size() >= a[1].size() && b[1].size() >= a[2].size()) {
		for (int v : b[0]) {
			if (a[1].empty()) {
				r[v] = a[0].back(); a[0].pop_back();
			} else {
				r[v] = a[1].back(); a[1].pop_back();
			}
		}
		for (int v : b[1]) {
			if (a[2].empty()) {
				r[v] = a[0].back(); a[0].pop_back();
			} else {
				r[v] = a[2].back(); a[2].pop_back();
			}
		}
	} else {
		int k;
		if (b[0].size() <= a[0].size()) k = 0;
		else if (b[1].size() <= a[0].size()) k = 1;
		else throw 1;
		for (int v : b[k]) {
			r[v] = a[0].back(); a[0].pop_back();
		}
		for (int v : b[k^1]) {
			for (int i = 0; i < 3; ++ i) if (!a[i].empty()) {
				r[v] = a[i].back(); a[i].pop_back();
				break;
			}
		}
	}
	for (auto x : r) cout << x << " ";
	cout << endl;
}