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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000000000000

void assign_all(vector<int> &ans,vector<int> &num,vector<int> &vertex){
	while(num.size()!=0&&vertex.size()!=0){
		ans[vertex.back()] = num.back();
		num.pop_back();
		vertex.pop_back();
	}
}

int main(){
	
	int N;
	cin>>N;
	
	vector<vector<int>> E(N,vector<int>());
	for(int i=0;i<N-1;i++){
		int a,b;
		cin>>a>>b;
		a--;b--;
		E[a].push_back(b);
		E[b].push_back(a);
	}
	
	vector<int> dis(N,1000);
	dis[0] = 0;
	queue<int> Q;
	Q.push(0);
	
	while(Q.size()!=0){
		int u = Q.front();
		Q.pop();
		for(int i=0;i<E[u].size();i++){
			int v = E[u][i];
			if(dis[v]!=1000)continue;
			dis[v] = dis[u]^1;
			Q.push(v);
		}
	}
	vector<vector<int>> num(3,vector<int>());
	vector<vector<int>> vertex(2,vector<int>());
	for(int i=1;i<=N;i++){
		num[i%3].push_back(i);
		vertex[dis[i-1]].push_back(i-1);
	}
	
	
	vector<int> ans(N,-1);
	
	if(vertex[0].size()<=num[0].size())assign_all(ans,num[0],vertex[0]);
	if(vertex[1].size()<=num[0].size())assign_all(ans,num[0],vertex[1]);
	assign_all(ans,num[1],vertex[0]);
	assign_all(ans,num[2],vertex[1]);
	
	for(int i=0;i<3;i++){
		for(int j=0;j<2;j++){
			assign_all(ans,num[i],vertex[j]);
		}
	}
	
	for(int i=0;i<N;i++){
		if(i!=0)cout<<' ';
		cout<<ans[i];
	}
	cout<<endl;
	
	
	return 0;
}