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;
#define SZ(x) ((int)(x).size())

typedef	long long ll;

const int MAXN = 2e5+1;

vector<int> G[MAXN];
vector<int> side[2];
vector<int> colors[3];
int choice[MAXN];

void dfs(int u, int p, int s)
{
	side[s].push_back(u);
	for (auto v : G[u])
		if (v != p)
			dfs(v,u,!s);
}

int main(void)
{
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	int N;
	cin >> N;
	for (int i(1); i < N; ++i)
	{
		int u, v;
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}

	dfs(1,1,0);
	for (int i(1); i <= N; ++i)
		colors[i%3].push_back(i);
	if (SZ(side[0]) > SZ(side[1]))
		swap(side[0], side[1]);

	if (SZ(side[0]) <= SZ(colors[0]))
	{
		while (SZ(side[0]) > 0)
		{
			choice[side[0].back()] = colors[0].back();
			colors[0].pop_back();
			side[0].pop_back();
		}
		for (int c(0); c < 3; ++c)
			while (SZ(colors[c]) > 0)
			{
				choice[side[1].back()] = colors[c].back();
				side[1].pop_back();
				colors[c].pop_back();
			}
	}
	else
	{
		while (SZ(colors[1]) > 0)
		{
			choice[side[1].back()] = colors[1].back();
			side[1].pop_back();
			colors[1].pop_back();
		}
		while (SZ(side[1]) > 0)
		{
			choice[side[1].back()] = colors[0].back();
			side[1].pop_back();
			colors[0].pop_back();
		}
		for (int c : {0, 2})
			while (SZ(colors[c]) > 0)
			{
				choice[side[0].back()] = colors[c].back();
				side[0].pop_back();
				colors[c].pop_back();
			}
	}

	for (int i(1); i <= N; ++i)
		cout << choice[i] << ' ';
	cout << endl;
}