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;

#define int long long

int a[100010];

int v[200010][4] = {};
int num[200010][4] = {};

vector<int> tree[200010];

void dfs(int idx, int cnt, int c, int old){
	if(cnt == 0){
		c ^= 1;
	}

	if(v[idx][cnt] == 1){
		if(cnt == 0 && num[idx][0] != c){
			cout << -1 << endl;
			exit(0);
		}
		return;
	}


	v[idx][cnt] = 1;
	num[idx][cnt] = c;

	for(int i = 0;i < tree[idx].size();i++){
		if(tree[idx][i] != old) dfs(tree[idx][i], (cnt+1)%3, c, idx);
	}

	return;
}

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

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

	queue<int> tama[3];
	for(int i = 1;i <= n;i++){
		tama[i%3].push(i);
	}

	int next = 0;
	for(int i = 0;i < n;i++){
		if(v[i][0] == 0){
			dfs(i, 0, next, 0);
			next ^= 1;
		}
	}

	vector<int> ans;
	for(int i = 0;i < n;i++){
		if(num[i][0] == 1){
			if(tama[1].size() > 0){
				ans.push_back(tama[1].front());
				tama[1].pop();
			}else if(tama[0].size() > 0){
				ans.push_back(tama[0].front());
				tama[0].pop();
			}else{
				cout << -1 << endl;
				return 0;
			}
		}else{
			if(tama[2].size() > 0){
				ans.push_back(tama[2].front());
				tama[2].pop(), -1;
			}else if(tama[0].size() > 0){
				ans.push_back(tama[0].front());
				tama[0].pop();
			}else{
				cout << -1 << endl;
				return 0;
			}
		}
	}

	for(int i = 0;i < ans.size();i++){
		cout << ans[i];
		if(i < ans.size()-1){
			cout << " ";
		}else{
			cout << endl;
		}
	}

	return 0;
}