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 n;
int u,v;
vector<int> ga[200100];
int vis[200100];
int mx[4];
int cnt[4];
int cz[3];
int res[200100];
void bipartite(int u,int co){//二分图染色 
	vis[u]=co;
	cz[co]++;
	for(int v:ga[u]){
		if(!vis[v]){
			bipartite(v,3-co);
		}
	}
}
void asgn(int vert,vector<int> ve){//把ve内容分配给vert，按序分配 
	for(int i=0;i<ve.size();i++){
		if(mx[ve[i]]<cnt[ve[i]]){
			res[vert]=3*mx[ve[i]]+ve[i];
			mx[ve[i]]++;
			return;
		}
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n;
	for(int i=1;i<n;i++){
		cin>>u>>v;
		ga[u].push_back(v);
		ga[v].push_back(u);
	} 
	bipartite(1,1);
	cnt[3]=n/3;
	cnt[1]=n/3;
	cnt[2]=n/3;
	if(n%3==1){
		cnt[1]++;
	}
	if(n%3==2){
		cnt[1]++;
		cnt[2]++;
	}
	if(cz[1]>cz[2]){
		swap(cz[1],cz[2]);
		for(int i=1;i<=n;i++){
			if(vis[i]==1)
				vis[i]=2;
			else
				vis[i]=1;
		}
	}
	if(cz[1]<=cnt[3]){
		for(int i=1;i<=n;i++){
			if(vis[i]==1)
				asgn(i,{3});
			else
				asgn(i,{1,2,3});
		} 
	}
	else{
		for(int i=1;i<=n;i++){
			if(vis[i]==1){
				asgn(i,{1,3});
			}
			else{
				asgn(i,{2,3});
			}
		} 
	}
	for(int i=1;i<=n;i++)
		cout<<res[i]<<' ';
	cout<<endl;
	return 0;
}