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>
#define maxn 200005
using namespace std;
int n;
vector<int> g[maxn];
int col[maxn],num[3];
int p[maxn],a[maxn];
void dfs(int u,int f,int c)
{
	col[u]=c;
	num[c]++;
	for(int v:g[u])if(v!=f)dfs(v,u,3-c);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,0,1);
	int x=1,y=2,z=3;
	if(num[1]<=n/3)
	{
		for(int i=1;i<=n;++i)if(col[i]==1)p[i]=z,z+=3;
		for(int i=1;i<=n;++i)if(!p[i])
		{
			if(x<=n)p[i]=x,x+=3;
			else if(y<=n)p[i]=y,y+=3;
			else p[i]=z,z+=3;
		}
	}
	else if(num[2]<=n/3)
	{
		for(int i=1;i<=n;++i)if(col[i]==2)p[i]=z,z+=3;
		for(int i=1;i<=n;++i)if(!p[i])
		{
			if(x<=n)p[i]=x,x+=3;
			else if(y<=n)p[i]=y,y+=3;
			else p[i]=z,z+=3;
		}
	}
	else
	{
		for(int i=1;i<=n;++i)
		{
			if(col[i]==1)
			{
				if(x<=n)p[i]=x,x+=3;
				else p[i]=z,z+=3;
			}
			else
			{
				if(y<=n)p[i]=y,y+=3;
				else p[i]=z,z+=3;
			}
		}
	}
	for(int i=1;i<=n;++i)printf("%d ",p[i]);
}