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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
using ll = long long;

const int N = 2 * (int)1e5;
vector<int> conns[N];
int side[N];
int cou[N];

void dfs(int i, int p) {
	side[i] = 1 ^ side[p];
	++cou[side[i]];
	for (auto t : conns[i]) if (t != p) dfs(t, i);
}

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

	int n;
	cin >> n;
	for (int i = 0; i < n-1; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		conns[a].push_back(b);
		conns[b].push_back(a);
	}
	dfs(0, 0);

	int t1 = (n+2) / 3;
	int s1 = (cou[1] > cou[0]);
	cou[s1] -= t1;

	int t2 = (n+1) / 3;
	int s2 = (cou[1] > cou[0]);
	cou[s2] -= t2;

	if (cou[s1] < 0 || cou[s2] < 0) cout << -1 << '\n';
	else {
		vector<int> inds = {3, 1, 2};
		for (int i = 0; i < n; ++i) {
			if (side[i] == s1 && inds[1] <= n) {
				cout << inds[1] << ' ';
				inds[1] += 3;
			} else if (side[i] == s2 && inds[2] <= n) {
				cout << inds[2] << ' ';
				inds[2] += 3;
			} else {
				cout << inds[0] << ' ';
				inds[0] += 3;
			}
		}
		cout << '\n';
	}
}