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>
#define N 201000
using namespace std;
vector <int> v[N], G[3], V[2];
int a[N], n, col[N], ch[N], res[N];
void dfs(int a, int b){
	col[a] = !col[b];
	V[col[a]].push_back(a);
	for(auto &x : v[a]){
		if(x == b) continue;
		dfs(x,a);
	}
}
int main() {
	int a, b;
	cin >> n;
	for(int i = 1; i<= n; i++) G[i%3].push_back(i);
	for(int i = 1; i< n; i++){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	dfs(1,0);
	int pv = 0;
	if(V[pv].size() > V[!pv].size()) pv = !pv;
	if(V[pv].size() <= G[0].size()){
		int pp = 0;
		for(auto &x: V[pv]){
			res[x] = G[0][pp];
			ch[G[0][pp]] = 1;
			pp++;
		}
	}
	else {
		int pp = 0, pp2 = 0;
		for(auto &x: V[pv]){
			if(pp < G[1].size()){
				res[x] = G[1][pp];
				ch[G[1][pp]] = 1;
				pp++;
			}
			else {
				res[x] = G[0][pp2];
				ch[G[0][pp2]] = 1;
				pp2++;
			}
		}
	}
	vector <int> p;
	for(int i=1; i<=n; i++) if(!ch[i]) p.push_back(i);
	for(int i=0; i<V[!pv].size(); i++) res[V[!pv][i]] = p[i];
	for (int i = 1; i <= n; i++)cout << res[i] << " ";
	return 0;
}