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<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=200005;
int n,i,j,k,head[N],adj[N*2],nxt[N*2],dep[N],p[N],q[N],cnt1,cnt2,id[N];
bool v[N];
void dfs(int x,int dad)
{
	if(dep[x]&1)
		p[++cnt1]=x;
	else
		q[++cnt2]=x;
	for(int y=head[x];y;y=nxt[y])
		if(adj[y]!=dad)
		{
			dep[adj[y]]=dep[x]+1;
			dfs(adj[y],x);
		}
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		adj[i*2-1]=k;
		nxt[i*2-1]=head[j];
		head[j]=i*2-1;
		adj[i*2]=j;
		nxt[i*2]=head[k];
		head[k]=i*2;
	}
	dfs(1,-1);
	if(cnt1>n/3&&cnt2>n/3)
	{
		for(i=1,j=1;i<=n;i+=3,++j)
			id[p[j]]=i;
		for(i=2,j=1;i<=n;i+=3,++j)
			id[q[j]]=i;
		for(i=1;i<=n;++i)
			v[id[i]]=true;
		for(i=1,j=1;i<=n;++i)
			if(!id[i])
			{
				while(v[j])
					++j;
				id[i]=j++;
			}
	}
	else
	{
		if(cnt1<cnt2)
		{
			for(i=3,j=1;j<=cnt1;i+=3,++j)
				id[p[j]]=i;
		}
		else
		{
			for(i=3,j=1;j<=cnt2;i+=3,++j)
				id[q[j]]=i;
		}
		for(i=1;i<=n;++i)
			v[id[i]]=true;
		for(i=1,j=1;i<=n;++i)
			if(!id[i])
			{
				while(v[j])
					++j;
				id[i]=j++;
			}
	}
	for(i=1;i<=n;++i)
		printf("%d ",id[i]);
	return 0;
}