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;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int n;
	cin >> n;
	vector<vector<int>> ad(n);
	for(int i=1;i<n;i++){
		int a,b;
		cin >> a >> b;
		a--,b--;
		ad[a].push_back(b);
		ad[b].push_back(a);
	}
	
	vector<int> res(n);
	vector<vector<int>> ta(2);
	function<void(int,int,int)> dfs=[&](int cur,int par,int dep){
		ta[dep&1].push_back(cur);
		for(int ch:ad[cur]){
			if(ch==par) continue;
			dfs(ch,cur,dep+1);
		}
	};
	dfs(0,-1,0);
	
	if(ta[0].size()<ta[1].size()) ta[0].swap(ta[1]);
	
	for(int i=1;i<=n;i+=3){
		int v=ta[0].back();
		res[v]=i;
		ta[0].pop_back();
	}
	
	if(ta[0].size()<ta[1].size()) ta[0].swap(ta[1]);
	
	for(int i=2;i<=n;i+=3){
		int v=ta[0].back();
		res[v]=i;
		ta[0].pop_back();
	}
	
	for(int i=3;i<=n;i+=3){
		if(ta[0].size()){
			int v=ta[0].back();
			res[v]=i;
			ta[0].pop_back();
		}else{
			int v=ta[1].back();
			res[v]=i;
			ta[1].pop_back();
		}
	}
	
	for(int i=0;i<n;i++){
		cout << res[i] << " ";
	}
	cout << "\n";
	return 0;
}