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=200005;
int n,a[N],ans[N];
vector<int>e[N],v[2];
void dfs(int x,int fa,int d)
{
	v[d].push_back(x);
	for(int to:e[x])
		if(to!=fa) 
			dfs(to,x,d^1);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0,0);
	int k1=n/3,k2=n/3;
	if(n%3==1) 
		k1++;
	else if(n%3==2) 
		k1++,k2++;
	if(v[0].size()>v[1].size()) 
		swap(v[0],v[1]);
	if(k1<=v[0].size()&&k2<=v[1].size())
	{
		for(int i=0;i<k1;i++) 
			ans[v[0][i]]=1;
		for(int i=0;i<k2;i++) 
			ans[v[1][i]]=2;
	}
	else
	{
		for(int i=0;i<k1;i++) 
			ans[v[1][i]]=1;
		for(int i=k1;i<k1+k2;i++) 
			ans[v[1][i]]=2;
	}
	int now0=3,now1=1,now2=2;
	for (int i=1;i<=n;i++)
	{
		if(!ans[i]) 
			a[i]=now0,now0+=3;
		else if(ans[i]==1) 
			a[i]=now1,now1+=3;
		else 
			a[i]=now2,now2+=3;
	}
	for(int i=1;i<=n;i++) 
		printf("%d ",a[i]);
	return 0;
}