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;
int i,i0,n,m,ans,a[200005],cnt0,cnt1,cnt2,siz[200005],msiz[200005],vis[200005],c1,c0;
vector<int>mp[200005];
void dfs(int now,int root,int dep)
{
	for(int x:mp[now])if(x!=root)dfs(x,now,dep+1);
	if(dep%2)
	{
		c1++;
		if(cnt1)a[now]=(cnt1-1)*3+1,cnt1--;
		else a[now]=cnt0*3,cnt0--;
	}
	else
	{
		c0++;
		if(cnt2)a[now]=(cnt2-1)*3+2,cnt2--;
		else a[now]=cnt0*3,cnt0--;
	}
}
void dfs0(int now,int root,int dep)
{
	for(int x:mp[now])if(x!=root)dfs0(x,now,dep+1);
	if(c1>c0)
	{
		if(dep%2)
		{
			if(cnt1)a[now]=(cnt1-1)*3+1,cnt1--;
			else if(cnt2) a[now]=(cnt2-1)*3+2,cnt2--;
			else a[now]=cnt0*3,cnt0--;
		}
		else
		{
			a[now]=cnt0*3,cnt0--;
		}
	}
	else
	{
		if(dep%2==0)
		{
			if(cnt1)a[now]=(cnt1-1)*3+1,cnt1--;
			else if(cnt2) a[now]=(cnt2-1)*3+2,cnt2--;
			else a[now]=cnt0*3,cnt0--;
		}
		else
		{
			a[now]=cnt0*3,cnt0--;
		}
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d %d",&x,&y);
		mp[x].push_back(y);
		mp[y].push_back(x);
	}
	cnt0=n/3;
	cnt1=n/3+(n%3!=0);
	cnt2=n/3+(n%3==2);
	dfs(1,-1,0);
	if(cnt0==0)for(int i=1;i<=n;i++)printf("%d%c",a[i],i==n?'\n':' ');
	else
	{
		cnt0=n/3;
		cnt1=n/3+(n%3!=0);
		cnt2=n/3+(n%3==2);
		dfs0(1,-1,0);
		for(int i=1;i<=n;i++)printf("%d%c",a[i],i==n?'\n':' ');
	}
	return 0;
}