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 M=1000000007;
const int N=1000005;
int i,n,head[N],Next[N],adj[N],p[N],vis[N],j,k,ans[N],u,v;
vector<int> e[2];
void Push(int u,int v)
{
	Next[++k]=head[u];
	head[u]=k;
	adj[k]=v;
}
void dfs(int i,int fa,int d)
{
	int j;
	e[d].push_back(i);
	for(j=head[i];j!=0;j=Next[j])
		if(adj[j]!=fa)
			dfs(adj[j],i,d^1);
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d %d",&u,&v);
		Push(u,v);
		Push(v,u);
	}
	dfs(1,0,0);
	if(e[0].size()>e[1].size())
		swap(e[0],e[1]);
	if(e[0].size()<=n/3)
	{
		for(i=3,j=0;i<=n&&j<e[0].size();i+=3,++j)
		{
			ans[e[0][j]]=i;
			vis[i]=1;
		}
	}
	else
	{
		for(i=1,j=0;i<=n&&j<e[0].size();i+=3,++j)
		{
			ans[e[0][j]]=i;
			vis[i]=1;
		}
		for(i=3;i<=n&&j<e[0].size();i+=3,++j)
		{
			ans[e[0][j]]=i;
			vis[i]=1;
		}
	}
	j=0;
	for(i=1;i<=n;++i)
		if(vis[i]==0)
			ans[e[1][j++]]=i;
	for(i=1;i<=n;++i)
		printf("%d ",ans[i]);
		
}