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>
#define pb push_back

using namespace std;
using ll = long long int;
const int MOD = 1e9 + 7;

ll C2(ll n)
{
	return n*(n-1)/2;
}

vector<vector<int>> part(2);

void dfs(vector<vector<int>> &g, int u, int v, int par)
{
	part[par].pb(u);
	for (auto x : g[u])
	{
		if (x == v) continue;
		dfs(g, x, u, par^1);
	}
}

int main()
{
	ios::sync_with_stdio(0); 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;
		g[--u].pb(--v);
		g[v].pb(u);
	}
	ll zero = n/3, one = (n+2)/3;
	ll two = n - zero - one;
	dfs(g, 0, 0, 0);
	if (part[0].size() < part[1].size()) swap(part[0], part[1]);
	vector<int> final(n, 0);
	if (part[1].size() < two)
	{
		if (part[0].size() < one + two) {cout << -1; return 0;}
		int ct = 0;
		for (auto u : part[0])
		{
			if (ct < one) final[u] = 1;
			else if (ct < one + two) final[u] = 2;
			++ct;
		}
	}
	else
	{
		int ct = 0;
		for (auto u : part[0])
		{
			if (ct < one) final[u] = 1;
			++ct;
		}
		ct = 0;
		for (auto u : part[1])
		{
			if (ct < two) final[u] = 2;
			++ct;
		}
	}
	int next[3] = {3, 1, 2};
	for (int i = 0; i < n; ++i)
	{
		cout << next[final[i]] << ' ';
		next[final[i]] += 3;
	}
}

// Have ax0, bx1, cx2
// At most C(a, 2) + ab + ac + bc