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;
using ll=long long;

int N,dist[200000],res[200000];
queue<int>que[3];
vector<int>v[200000];

void dfs(int x,int from){
	for(int i:v[x]){
		if(i==from)continue;
		dist[i]=dist[x]+1;
		dfs(i,x);
	}
}

int main(){
	cin>>N;
	for(int i=1;i<N;i++){
		int a,b;
		cin>>a>>b;
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	dfs(0,0);
	for(int i=1;i<=N;i++){
		que[i%3].push(i);
	}
	vector<int>v[2];
	for(int i=0;i<N;i++){
		v[dist[i]&1].push_back(i);
	}
	if(v[0].size()<=N/3){
		for(int i:v[0]){
			res[i]=que[0].front();
			que[0].pop();
		}
		for(int i:v[1]){
			for(int j=0;j<3;j++){
				if(!que[j].empty()){
					res[i]=que[j].front();
					que[j].pop();
					break;
				}
			}
		}
	}
	else if(v[1].size()<=N/3){
		for(int i:v[1]){
			res[i]=que[0].front();
			que[0].pop();
		}
		for(int i:v[0]){
			for(int j=0;j<3;j++){
				if(!que[j].empty()){
					res[i]=que[j].front();
					que[j].pop();
					break;
				}
			}
		}
	}
	else{
		for(int i:v[0]){
			if(!que[1].empty()){
				res[i]=que[1].front();
				que[1].pop();
			}
			else{
				res[i]=que[0].front();
				que[0].pop();
			}
		}
		for(int i:v[1]){
			if(!que[2].empty()){
				res[i]=que[2].front();
				que[2].pop();
			}
			else{
				res[i]=que[0].front();
				que[0].pop();
			}
		}
	}
	for(int i=0;i<N;i++){
		cout<<res[i]<<' ';
	}
	cout<<endl;
}