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: Accepted)

//hitachi2020_c.cpp
//Sun Mar  8 20:53:12 2020

#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <math.h>
#include <set>
#define INTINF 2147483647
#define LLINF 9223372036854775807
#define MOD 1000000007

using namespace std;
using ll=long long;
typedef pair<int,int> P;

int main(){
	int n;
	cin >> n;

	vector<int> node[n];
	for (int i=0;i<n-1;i++){
		int a,b;
		cin >> a >> b;
		a--;b--;
		node[a].push_back(b);
		node[b].push_back(a);
	}

	int height[n];
	fill(height,height+n,-1);
	height[0] = 0;
	int heighttotal[2];
	fill(heighttotal,heighttotal+2,0);
	heighttotal[0] = 1;
	queue<int> q;
	q.push(0);
	while (q.size()){
		int temp = q.front();q.pop();
		for (int i=0;i<node[temp].size();i++){
			int childcand = node[temp][i];
			if (height[childcand]==-1){
				height[childcand] = height[temp]+1;
				heighttotal[height[childcand]%2]++;
				q.push(childcand);
			}
		}
	}

	int amari0=n/3;

	int ans[n];
	fill(ans,ans+n,-1);
	int cur1 = 1;
	int cur2 = 2;
	int cur0 = 3;
	if (heighttotal[0]>amari0 && heighttotal[1]>amari0){
		for (int i=0;i<n;i++){
			if(height[i]%2==0){
				if (cur1<=n){
					ans[i]=cur1;
					cur1 = cur1+3;
				}else {
					ans[i]=cur0;
					cur0 = cur0+3;
				}
			}else {
				if (cur2<=n){
					ans[i]=cur2;
					cur2 = cur2+3;
				}else {
					ans[i]=cur0;
					cur0 = cur0+3;
				}
			}
		}
	}else if (heighttotal[0]<=n/3){
		for (int i=0;i<n;i++){
			if(height[i]%2==0){
				ans[i]=cur0;
				cur0 = cur0+3;
			}else {
				if (cur2<=n){
					ans[i]=cur2;
					cur2 = cur2+3;
				}else if (cur1<=n){
					ans[i]=cur1;
					cur1 = cur1+3;
				}else {
					ans[i]=cur0;
					cur0 = cur0+3;
				}
			}
		}
	}else{
		for (int i=0;i<n;i++){
			if(height[i]%2==1){
				ans[i]=cur0;
				cur0 = cur0+3;
			}else {
				if (cur2<=n){
					ans[i]=cur2;
					cur2 = cur2+3;
				}else if (cur1<=n){
					ans[i]=cur1;
					cur1 = cur1+3;
				}else {
					ans[i]=cur0;
					cur0 = cur0+3;
				}
			}
		}
	}

	for (int i=0;i<n;i++){
		cout << ans[i] << " ";
	}
	printf("\n");

}