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;
typedef long long LL;
typedef vector<int> VI;
const int N=200200;
VI es[N];
int col[N];
VI cc[2];
int ans[N];
void dfs(int x,int fa=-1,bool isB=true){
	col[x]=isB?0:1;
	cc[col[x]].push_back(x);
	for(VI::iterator it=es[x].begin();it!=es[x].end();it++){
		if(*it!=fa){
			dfs(*it,x,!isB);
		}
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		es[u].push_back(v);
		es[v].push_back(u);
	}
	dfs(1);
	unsigned c1=(n+2)/3;
	unsigned c2=(n+1)/3;
	unsigned c3=(n+0)/3;
	if(cc[0].size()>cc[1].size()){
		swap(cc[0],cc[1]);
	}
	if(cc[0].size()>=c1&&cc[1].size()>=c2){
		for(unsigned i=0;i<c1;i++){
			ans[cc[0].back()]=i*3+1;
			cc[0].pop_back();
		}
		for(unsigned i=0;i<c2;i++){
			ans[cc[1].back()]=i*3+2;
			cc[1].pop_back();
		}
		for(unsigned i=0;i<c3;i++){
			if(!cc[0].empty()){
				ans[cc[0].back()]=i*3+3;
				cc[0].pop_back();
			}
			else{
				ans[cc[1].back()]=i*3+3;
				cc[1].pop_back();
			}
		}
	}
	else{
		assert(cc[0].size()<=c3);
		for(unsigned i=0;i<c3;i++){
			if(!cc[0].empty()){
				ans[cc[0].back()]=i*3+3;
				cc[0].pop_back();
			}
			else{
				ans[cc[1].back()]=i*3+3;
				cc[1].pop_back();
			}
		}
		for(unsigned i=0;i<c1;i++){
			ans[cc[1].back()]=i*3+1;
			cc[1].pop_back();
		}
		for(unsigned i=0;i<c2;i++){
			ans[cc[1].back()]=i*3+2;
			cc[1].pop_back();
		}
	}
	for(int i=1;i<=n;i++){
		printf("%d ",ans[i]);
	}
	printf("\n");
	return 0;
}