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 ll long long
#define ull unsigned long long
#define lsb(x) (x & (-x))

using namespace std;

vector <vector <int>> g, ids;

void dfs(int nod, int par, int e) {
	ids[e].push_back(nod);
	for(auto it : g[nod]) {
		if(it != par) {
			dfs(it, nod, e ^ 1);
		}
	}
}

int main() {
#ifdef HOME
	ifstream cin("C.in");
	ofstream cout("C.out");
#endif
	int i, n;
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n;
	g.resize(n + 1);
	for(i = 1; i < n; i++) {
		int x, y; cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}

	ids.resize(2);
	dfs(1, 0, 0);
	if(ids[0].size() > ids[1].size()) {
		swap(ids[0], ids[1]);
	}

	vector <int> cnt(3), sol(n + 1);
	for(i = 1; i <= n; i++) {
		cnt[i % 3]++;
	}

	if(ids[0].size() <= cnt[0]) {
		for(i = 0; i < ids[0].size(); i++) {
			sol[ids[0][i]] = 3 * cnt[0];
			cnt[0]--;
		}
		
	}
	else {
		for(i = 0; i < ids[0].size(); i++) {
			if(cnt[2] > 0) {
				cnt[2]--;
				sol[ids[0][i]] = 3 * cnt[2] + 2;
			}
			else {
				sol[ids[0][i]] = 3 * cnt[0];
				cnt[0]--;
			}
		}
	}
	for(i = 1; i <= n; i++) {
		if(sol[i]) continue;
		for(int j = 0; j < 3; j++) {
			if(cnt[j]) {
				sol[i] = j + (cnt[j] - (j > 0)) * 3;
				cnt[j]--;
				break;
			}
		}
	}

	for(i = 1; i <= n; i++) {
		cout << sol[i] << " ";
	}

	return 0;
}