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;

int main() {
    int N; cin >> N;
    vector<vector<int>> G(N);
    for(int i = 0; i < N - 1; ++i) {
        int a, b; cin >> a >> b;
        a--, b--;
        G[a].emplace_back(b);
        G[b].emplace_back(a);
    }
    vector<vector<int>> V(2);
    auto dfs = [&](auto &&dfs, int cur, int pre, int col) -> void {
        V[col].push_back(cur);
        for(auto &to : G[cur]) {
            if(to != pre) dfs(dfs, to, cur, 1 - col);
        }
    };
    dfs(dfs, 0, 0, 0);
    if(V[0].size() > V[1].size()) swap(V[0], V[1]);
    vector<int> ans(N);
    if(V[0].size() < N / 3) {
        int n = 3;
        for(auto &v : V[0]) {
            ans[v] = n;
            n += 3;
        }
        n = 1;
        for(auto &v : V[1]) {
            ans[v] = n++;
            if(n % 3 == 0) n++;
        }
    } else {
        int n = 1;
        for(auto &v : V[0]) {
            ans[v] = n;
            n += 3;
        }
        n = 2;
        for(auto &v : V[1]) {
            ans[v] = n++;
            if(n % 3 == 1) n++;
        }
    }
    for(auto &a : ans) cout << a << " ";
    cout << '\n';
    return 0;
}