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;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
vector<int> e[200005];
int n,s[200005],a[2];
void dfs(int x,int y,int z){
	s[x]=z,a[z]++;
	for(int i=0;i<e[x].size();i++)
		if(e[x][i]!=y)dfs(e[x][i],x,1-z);
}
int main(){
	n=read();
	for(int i=1;i<n;i++){
		int x=read(),y=read();
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0,0);
	int n0=n/3,n1=n/3+((n%3)>0),n2=n-n1-n0;
	for(int j=0;j<2;j++)
		if(a[j]<=n0){
			for(int i=1;i<=n;i++)
				if(s[i]==j){
					printf("%d ",n0*3);
					n0--;
				}
				else{
					if(n1){
						printf("%d ",n1*3-2);
						n1--;
					}
					else if(n2){
						printf("%d ",n2*3-1);
						n2--;
					}
					else{
						printf("%d ",n0*3);
						n0--;
					}
				}
			return 0;
		}
	for(int i=1;i<=n;i++)
		if(s[i]==0){
			if(n1){
				printf("%d ",n1*3-2);
				n1--;
			}
			else{
				printf("%d ",n0*3);
				n0--;
			}
		}
		else{
			if(n2){
				printf("%d ",n2*3-1);
				n2--;
			}
			else{
				printf("%d ",n0*3);
				n0--;
			}
		}
	return 0;
}