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<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int x,y,n,pr[200005];
vector<int>s[200005],deg[3],degp[2];
void dfs(int p,int f,int st)
{
	int v;
	degp[st].push_back(p);
	for(int i=0;i<s[p].size();i++)
	{
		v=s[p][i];
		if(v==f)continue;
		dfs(v,p,!st);
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)deg[i%3].push_back(i);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		s[x].push_back(y);
		s[y].push_back(x);
	}
	dfs(1,0,0);
	if(degp[0].size()>degp[1].size())swap(degp[0],degp[1]);
	if(degp[0].size()<=deg[0].size())
	{
		for(int i=0;i<degp[0].size();i++)
		{
			x=degp[0][i];
			pr[x]=deg[0].back();
			deg[0].pop_back();
		}
		for(int i=0;i<degp[1].size();i++)
		{
			x=degp[1][i];
			if(deg[0].size()>0)
			{
				pr[x]=deg[0].back();
			deg[0].pop_back();
			}
			else
			{
				if(deg[1].size()>0)
				{
					pr[x]=deg[1].back();
			deg[1].pop_back();
				}
				else
				{
					pr[x]=deg[2].back();
			deg[2].pop_back();
				}
			}
		}
	}
	else
	{
		if(deg[1].size()<deg[2].size())swap(deg[1],deg[2]);
		for(int i=0;i<degp[0].size();i++)
		{
			x=degp[0][i];
			if(deg[1].size()>0)
			{
			pr[x]=deg[1].back();
			deg[1].pop_back();
			}
			else
			{
					pr[x]=deg[0].back();
			deg[0].pop_back();
			}
		}
		for(int i=0;i<degp[1].size();i++)
		{
			x=degp[1][i];
			if(deg[0].size()>0)
			{
				pr[x]=deg[0].back();
			deg[0].pop_back();
			}
			else
			{
					pr[x]=deg[2].back();
			deg[2].pop_back();
			}
		}
	}
	for(int i=1;i<=n;i++)printf("%d ",pr[i]);
	return 0;	
}