Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> edge[200005];
ll a[200005]={0};
ll b[200006]={0};
vector<ll> zhe;
vector<ll> fu;
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++)
	{
		printf("%lld ",a[i]);
	}
	printf("\n");
	for(ll i=1;i<=n;i++)
	{
		if(a[i]==1)
		{
			zhe.push_back(i);
		}
		else
		{
			fu.push_back(i);
		}
	}
	if(zhe.size()>fu.size())
	{
		swap(zhe,fu);
	}
	if(zhe.size()<=n/3)
	{
		ll c=0;
		for(ll i=1;c<zhe.size();i++)
		{
			if(i%3==0)
			{
				b[i]=zhe[c++];
			}
		}
		c=0;
		for(ll i=1;i<=n;i++)
		{
			if(b[i]==0)
			{
				b[i]=fu[c++];
			}
		}
	}
	else
	{
		ll c=0;
		for(ll i=1;i<=n;i++)
		{
			if(i%3==1)
			{
				b[i]=zhe[c++];
			}
		}
		for(ll i=1;c<zhe.size();i++)
		{
			if(i%3==0)
			{
				b[i]=zhe[c++];
			}
		}
		c=0;
		for(ll i=1;i<=n;i++)
		{
			if(b[i]==0)
			{
				b[i]=fu[c++];
			}
		}
	}
	for(ll i=1;i<=n;i++)
	{
		printf("%lld ",b[i]);
	}
}