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;
#define vi vector<int>
#define vll vector<ll>
#define pll pair<ll,ll>
#define pii pair<int,int>
#define pb  push_back
#define ppb pop_back
#define mp  make_pair
#define SIZE 200005
vector<ll> graph[SIZE];
ll ans[SIZE];
set<ll> A;
set<ll> B;
void bipartite(ll u,ll p,ll color){
	if(color) B.insert(u);
	else A.insert(u);
	// cout<<u<<endl;
	for(auto v:graph[u]){
		if(v!=p)
			bipartite(v,u,1-color);
	}
}
int main(){
	// #ifndef ONLINE_JUDGE
	// freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);
	// #endif
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);
	ll N;
	cin>>N;
	for(ll i=0;i<N-1;++i){
		ll u,v;
		cin>>u>>v;
		graph[u].pb(v);
		graph[v].pb(u);
	}
	A.clear();
	B.clear();
	bipartite(1,0,0);
	if(A.size()>B.size()){
		swap(A,B);
	}
	// for(auto e:A){
	// 	cout<<e<<" ";
	// }
	// cout<<endl;
	// for(auto e:B){
	// 	cout<<e<<" ";
	// }
	// cout<<endl;
	if(A.size()<=(N/3)){
		ll first=3;
		for(auto e:A){
			ans[e]=first;
			first=first+3;
		}
		first=1;
		for(auto e:B){
			if(first%3==0){
				first++;
			}
			ans[e]=first;
			first++;
		}
		for(ll i=0;i<N;++i){
			cout<<ans[i+1]<<" ";
		}
	}
	else{
		ll first=1;
		for(auto e:A){
			if(first>N){
				first=3;
			}
			ans[e]=first;
			first=first+3;
		}
		if(first>N) first=3;
		for(auto e:B){
			if(first>N) first=2;
			ans[e]=first;
			first+=3;
		}
		for(ll i=0;i<N;++i){
			cout<<ans[i+1]<<" ";
		}
	}
}