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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <list>
#include <algorithm>
#include <cmath>

using namespace std;

static const int MAX = 200000;

int n;
vector<int> M[MAX];
vector<int> K[MAX];

void bfs(int s)
{
    int d[MAX] = {};
    queue<int> q;
    q.push(s);
    d[s] = 0;
    int u;
    while (!q.empty()) {
        u = q.front(); q.pop();
        for (int i = 0; i < M[u].size(); i++) {
            int v = M[u][i];
            if (d[v] != 0) continue;
            d[v] = d[u] + 1;
            if (d[v] < 3) {
                q.push(v);
            } else {
                K[s].push_back(v);
            }
        }
    }
}

int main()
{
    int i, j;
    int a, b;
    vector<int> all;

    cin >> n;

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

    for (i = 0; i < n; i++) {
        bfs(i);
        all.push_back(i + 1);
    }

    do {
        bool flg = true;
        for (i = 0; i < n; i++) {
            for (j = 0; j < K[i].size(); j++) {
                int v = K[i][j];
                if (((all[i] + all[v]) % 3 != 0) && ((all[i] * all[v]) % 3 != 0)) {
                    flg = false;
                    break;
                }
            }
            if (!flg) break;
        }
        if (flg) {
            for (auto itr = all.begin(); itr < all.end(); itr++) {
                cout << *itr << " ";
            }
            cout << endl;
            return 0;
        }
    } while (next_permutation(all.begin(), all.end()));

    cout << "-1" << endl;

    return 0;
}