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;
int n,ans[202020],ct1,ct2;
vector<int>v[202020],md[4],vv,cr,cl[3];
void dfs(int x,int bf,int cll)
{
	cl[cll].push_back(x);
	if(cll==0)
	{
		ct1++;
	}
	else
	{
		ct2++;
	}
	for(int i=0;i<v[x].size();i++)
	{
		int y=v[x][i];
		if(y==bf)
		{
			continue;
		}
		dfs(y,x,cll^1);
	}
}
int main()
{
	cin>>n;
	for(int i=0;i<n-1;i++)
	{
		int x,y;
		cin>>x>>y;
		x--;
		y--;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(0,-1,0);
	if(ct1>ct2)
	{
		swap(ct1,ct2);
		swap(cl[0],cl[1]);
	}
	for(int i=1;i<=n;i++)
	{
		md[i%3].push_back(i);
	}
	for(int i=0;i<cl[0].size();i++)
	{
		vv.push_back(cl[0][i]);
	}
	for(int i=0;i<cl[1].size();i++)
	{
		vv.push_back(cl[1][i]);
	}
	if(ct1>=md[2].size())
	{
		for(int i=0;i<md[2].size();i++)
		{
			cr.push_back(md[2][i]);
		}
		for(int i=0;i<md[0].size();i++)
		{
			cr.push_back(md[0][i]);
		}
		for(int i=0;i<md[1].size();i++)
		{
			cr.push_back(md[1][i]);
		}
	}
	else
	{
		for(int i=0;i<md[0].size();i++)
		{
			cr.push_back(md[0][i]);
		}
		for(int i=0;i<md[1].size();i++)
		{
			cr.push_back(md[1][i]);
		}
		for(int i=0;i<md[2].size();i++)
		{
			cr.push_back(md[2][i]);
		}
	}
	for(int i=0;i<n;i++)
	{
		ans[vv[i]]=cr[i];
	}
	for(int i=0;i<n;i++)
	{
		cout<<ans[i]<<' ';
	}
	cout<<endl;
	return 0;
}