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; 
ll n;
ll head[500005],to[500005],nxt[500005],cnt;
int c[500005];
bool ok;
int w,b,cw,cb;
inline void add(int x,int y){
	to[++cnt]=y;
	nxt[cnt]=head[x];
	head[x]=cnt;
}
inline void clr(int x,int cl){
	if(cl){
		w++;
	}
	else{
		b++;
	}
	c[x]=cl;
	vector<int> v;
	for(int i=head[x];i;i=nxt[i]){
		int y=to[i];
		if(c[y]!=-1){
			continue;
		}
		clr(y,cl^1);
	}
}
int main(){
	cin>>n;
	for(int i=1;i<n;++i){
		int a,b;
		cin>>a>>b;
		add(a,b); 
		add(b,a);
	}
	memset(c,-1,sizeof(c));
	clr(1,0);
	int rd=n/3;
	if(w>b){
		swap(w,b);
		cw=0,cb=1;
	}
	else{
		cw=1,cb=0;
	}
	if(w<=rd){
		int st=3,nd=1;
		for(int i=1;i<=n;++i){
			if(c[i]==cw){
				cout<<st<<' ';
				st+=3;
			}
			else{
				if(nd<=n){
					cout<<nd<<' ';
					nd++;
					if(nd%3==0){
						nd++;
					}
				}
				else{
					cout<<st<<' ';
					st+=3;
				}
			}
		}
	}
	else{
		int st=3,nd=1,rd=2;
		for(int i=1;i<=n;++i){
			if(c[i]==cw){
				if(nd<=n){
					cout<<nd<<' ';
					nd+=3;
				}
				else{
					cout<<st<<' ';
					st+=3;
				}
			}
			else{
				if(rd<=n){
					cout<<rd<<' ';
					rd+=3;
				}
				else{
					cout<<st<<' ';
					st+=3;
				}
			}
		}
	}
	cout<<'\n';
	return 0;
}