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;

vector<int> e[200010];
int s[200010], ns[2];

void dfs(int x, int p, int d) {
	s[x] = d;
	ns[d]++;
	for (auto y : e[x]) if (y != p) dfs(y,x,1-d);
}

int main() {
	ios::sync_with_stdio(false);
	int n; cin >> n;
	for (int i = 1; i < n; i++) {
		int a,b; cin >> a >> b;
		e[a].push_back(b);
		e[b].push_back(a);
	}
	dfs(1,0,0);
	int n0 = n/3, n1 = n/3+((n%3)>0), n2 = n-n1-n0;
	for (int j = 0; j < 2; j++) if (ns[j] <= n0) {
		for (int i = 1; i <= n; i++) {
			if (s[i] == j) { cout << n0*3 << " "; n0--; }
			else {
				if (n1) { cout << n1*3-2 << " "; n1--; }
				else if (n2) { cout << n2*3-1 << " "; n2--;	}
				else { cout << n0*3 << " "; n0--; }
			}
		}
		cout << "\n";
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		if (s[i] == 0) {
			if (n1) { cout << n1*3-2 << " "; n1--; }
			else { cout << n0*3 << " "; n0--; }
		}
		else {
			if (n2) { cout << n2*3-1 << " "; n2--; }
			else { cout << n0*3 << " "; n0--; }
		}
	}
	cout << "\n";
}