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:
// C - ThREE
#include <bits/stdc++.h>

using namespace std;

void dfs(const vector<vector<int> > &g, int root, int d, vector<int> &c)
{
	c[root] = d;
	for (int i = 0; i < g[root].size(); i++)
		if (c[g[root][i]] < 0) dfs(g, g[root][i], d ^ 1, c);
}

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0);
	int n;
	cin >> n;
	vector<vector<int> > g(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);
	}

	vector<int> ans(n), c(n, -1), a[2];
	dfs(g, 0, 0, c);
	for (int i = 0; i < n; i++) a[c[i]].push_back(i);
	set<int> s;
	for (int i = 1; i <= n; i++) s.insert(i);
	int x = a[0].size() < a[1].size() ? 0 : 1;
	if (a[x].size() <= n / 3) {
		for (int i = 0, next = 3; i < a[x].size(); i++, next += 3) {
			ans[a[x][i]] = next;
			s.erase(next);
		}
	} else {
		for (int i = 0, next = 1; i < a[x].size(); i++, next += 3) {
			ans[a[x][i]] = next;
			s.erase(next);
			if (next + 3 > n) next = 0;
		}
	}

	set<int>::iterator it = s.begin();
	for (int i = 0; i < n; i++)
		if (ans[i] == 0) ans[i] = *it++;

	for (int i = 0; i < n; i++) cout << (i ? " " : "") << ans[i];
	cout << endl;

	return 0;
}