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;
const int N=2e5+5;
vector<int> black,white;
vector<int> ch[3],e[N];
int dep[N],ans[N];
void dfs(int x,int fath){
	dep[x]=dep[fath]+1;
	if(dep[x]&1) black.push_back(x);
	else white.push_back(x);
	for(int i=0;i<e[x].size();++i)
		if(e[x][i]!=fath) dfs(e[x][i],x);
}
int main(){
	int n,x,y;cin>>n;
	for(int i=1;i<n;++i)
		cin>>x>>y,e[x].push_back(y),e[y].push_back(x);
	dfs(1,0);
	int sz=(n/3);
	if(black.size()>white.size()) swap(black,white);
	if(black.size()<=sz){
		int now=3;
		for(int i=0;i<black.size();++i)
			ans[black[i]]=now,now+=3;
		int T=0;if(now>n) now=++T;
		for(int i=0;i<white.size();++i){
			ans[white[i]]=now,now+=3;
			if(now>n) now=++T;
		}
	}
	else{
		int now=1;
		for(int i=0;i<black.size();++i){
			ans[black[i]]=now,now+=3;
			if(now>n) now=3;
		}
		int T=now;now=2;
		for(int i=0;i<white.size();++i){
			ans[white[i]]=now,now+=3;
			if(now>n) now=T;
		}
	}
	for(int i=1;i<=n;++i)
		cout<<ans[i]<<" ";
	return 0;
}