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 sz(a) ((int)(a.size()))
#define sg sz(ve[g])
using namespace std;

const int N = 200005;
int n, x, y, fp[N];
vector<int> v[N], ve[2];
bool g, gr[N];

void sgd(int dg, int op, bool bo) {
	ve[bo].push_back(dg);
	for(int i = 0; i < v[dg].size(); ++i) {
		if(v[dg][i] != op) {
			sgd(v[dg][i], dg, bo ^ 1);
		}
	}
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for(int i = 1; i < n; ++i) {
		cin >> x >> y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	sgd(1, 0, 0);
	if(sz(ve[0]) > sz(ve[1])) g = 1;
	if(n / 3 >= sg) {
		int j3 = 3;
		for(int i = 0; i < ve[g].size(); ++i) {
			fp[ve[g][i]] = j3;
			gr[j3] = 1;
			j3 += 3;
		}
	} else {
		int j1 = 1, j2 = 2;
		for(int i = 0; j1 <= n; ++i) {
			fp[ve[0][i]] = j1;
			gr[j1] = 1;
			j1 += 3;
		}
		for(int i = 0; j2 <= n; ++i) {
			fp[ve[1][i]] = j2;
			gr[j2] = 1;
			j2 += 3;
		}
	}
	int dgr = 1;
	for(int i = 1; i <= n; ++i) {
		while(gr[dgr]) ++dgr;
		if(fp[i] == 0) {
			fp[i] = dgr;
			gr[dgr] = 1;
		}
	}
	for(int i = 1; i <= n; ++i) {
		cout << fp[i] << " ";
	}
	return 0;
}