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;

inline int read()
{
	int sum=0,ff=1; char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-') ff=-1;
		ch=getchar();
	}
	while(isdigit(ch))
		sum=sum*10+(ch^48),ch=getchar();
	return sum*ff;
}

const int N=2e5+5;

int n,m,cnt,ans[N],head[N];
struct nood
{
	int nex,to;
};
nood e[N<<1];

vector<int> col[2],mo[4];

inline void jia(int u,int v)
{
	e[++cnt].nex=head[u];
	head[u]=cnt;
	e[cnt].to=v;
}

inline void dfs(int u,int fa,int co)
{
	col[co].push_back(u);
	for ( int i=head[u];i;i=e[i].nex ) 
		if(e[i].to!=fa) dfs(e[i].to,u,co^1);
}

inline void getans(int co,int mod) 
{
	while(!col[co].empty()&&!mo[mod].empty())
	{
		ans[col[co].back()]=mo[mod].back();
		col[co].pop_back();
		mo[mod].pop_back();
	}
}

int main()
{
	n=read();
	for ( int i=1;i<=n;i++ ) mo[i%3].push_back(i);
	for ( int i=1;i<n;i++ )
	{
		int u,v;
		u=read(),v=read();
		jia(u,v);
		jia(v,u);
	}
	dfs(1,0,0);
	getans(col[0].size()<col[1].size(),1);
	getans(col[0].size()<col[1].size(),2);
	getans(1,0);
	getans(0,0);
	for ( int i=1;i<=n;i++ ) printf("%d ",ans[i]);
	return 0;
}