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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

int N;
vector<int> graph[200005];
int par[200005];
int clr[200005];
vector<int> lst[3];
int ans[200005];
int cnt[2];

void dfs(int n){
	for(int e : graph[n]){
		if(e != par[n]){
			par[e] = n;
			clr[e] = clr[n] ^ 1;
			dfs(e);
			cnt[clr[e]]++;
		}
	}
}

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> N;
	for(int i = 1; i<N; i++){
		int a, b;
		cin >> a >> b;
		graph[a].push_back(b);
		graph[b].push_back(a);
	}
	dfs(1);
	for(int i = 1; i<=N; i++){
		lst[i%3].push_back(i);
	}
	if(cnt[0] < cnt[1]){
		swap(lst[1], lst[2]);
	}
	if(cnt[0] <= N/3){
		for(int i = 1; i<=N; i++){
			if(clr[i] == 0){
				ans[i] = lst[0].back();
				lst[0].pop_back();
			}
			else if(lst[1].size()){
				ans[i] = lst[1].back();
				lst[1].pop_back();
			}
			else if(lst[2].size()){
				ans[i] = lst[2].back();
				lst[2].pop_back();
			}
			else{
				ans[i] = lst[0].back();
				lst[0].pop_back();
			}
		}
	}
	else if(cnt[1] <= N/3){
		for(int i = 1; i<=N; i++){
			if(clr[i] == 1){
				ans[i] = lst[0].back();
				lst[0].pop_back();
			}
			else if(lst[1].size()){
				ans[i] = lst[1].back();
				lst[1].pop_back();
			}
			else if(lst[2].size()){
				ans[i] = lst[2].back();
				lst[2].pop_back();
			}
			else{
				ans[i] = lst[0].back();
				lst[0].pop_back();
			}
		}
	}
	else{
		for(int i = 1; i<=N; i++){
			if(!clr[i]){
				if(lst[1].size()){
					ans[i] = lst[1].back();
					lst[1].pop_back();
				}
				else if(lst[0].size()){
					ans[i] = lst[0].back();
					lst[0].pop_back();
				}
				else{
					while(1);
				}
			}
			else{
				if(lst[2].size()){
					ans[i] = lst[2].back();
					lst[2].pop_back();
				}
				else if(lst[0].size()){
					ans[i] = lst[0].back();
					lst[0].pop_back();
				}
				else{
					while(1);
				}			
			}
		}
	}
	for(int i =1 ; i<=N; i++){
		cout << ans[i] << " ";
	}
}