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;
typedef long long ll;
vector<ll> edge[200005];
ll a[200005]={0};
ll b[200006]={0};
ll b1[200006]={0};
vector<ll> positive;
vector<ll> negative;
int main(void)
{
	ll n;
	scanf("%lld",&n);
	for(ll i=0;i<n-1;i++)
	{
		ll from,to;
		scanf("%lld%lld",&from,&to);
		edge[from].push_back(to);
		edge[to].push_back(from);
	}
	queue<ll> que;
	que.push(1);
	a[1]=1;
	while(que.size()!=0)
	{
		ll q=que.front();
		que.pop();
		ll col=a[q];
		for(ll i=0;i<edge[q].size();i++)
		{
			if(a[edge[q][i]]==0)
			{
				a[edge[q][i]]=col*-1;
				que.push(edge[q][i]);
			}
		}
	}
	for(ll i=1;i<=n;i++)
	{
		if(a[i]==1)
		{
			positive.push_back(i);
		}
		else
		{
			negative.push_back(i);
		}
	}
	if(positive.size()>negative.size())
	{
		swap(positive,negative);
	}
	if(positive.size()<=n/3)
	{
		ll c=0;
		for(ll i=1;i<=n;i++)
		{
			if(i%3==0&&c<positive.size())
			{
				b[i]=positive[c++];
			}
		}
		c=0;
		for(ll i=1;i<=n;i++)
		{
			if(b[i]==0)
			{
				b[i]=negative[c++];
			}
		}
	}
	else
	{
		ll c=0;
		for(ll i=1;i<=n;i++)
		{
			if(i%3==1&&c<positive.size())
			{
				b[i]=positive[c++];
			}
		}
		for(ll i=1;i<=n;i++)
		{
			if(i%3==0&&c<positive.size())
			{
				b[i]=positive[c++];
			}
		}
		c=0;
		for(ll i=1;i<=n;i++)
		{
			if(b[i]==0)
			{
				b[i]=negative[c++];
			}
		}
	}
	for(ll i=1;i<=n;i++)
	{
		b1[b[i]]=i;
	}
	for(ll i=1;i<=n;i++)
	{
		printf("%lld ",b1[i]);
	}
}