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)

#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].push_back(b);
    g[b].push_back(a);
  }

  vector<vector<int>> c(2);
  auto dfs = [&](auto&& dfs, int i, int j, int d) -> void {
    c[d % 2].push_back(j);
    for (auto k : g[j]) {
      if (k == i) continue;
      dfs(dfs, j, k, d + 1);
    }
  };
  dfs(dfs, 0, 0, 0);
  if (c[0].size() > c[1].size()) swap(c[0], c[1]);

  vector<vector<int>> p(3);
  if (c[0].size() <= n / 3) {
    for (int i = 1; i <= n; i++) {
      if (i % 3 == 0) {
        if (!c[0].empty()) {
          p[0].push_back(c[0].back());
          c[0].pop_back();
        } else {
          // if (c[1].empty()) return 1;
          p[0].push_back(c[1].back());
          c[1].pop_back();
        }
      } else {
        p[i % 3].push_back(c[1].back());
        c[1].pop_back();
      }
    }
  } else {
    for (int i = 1; i <= n; i++) {
      if (i % 3 == 1) {
        // if (c[1].empty()) return 1;
        p[1].push_back(c[1].back());
        c[1].pop_back();
      } else if (i % 3 == 2) {
        // if (c[0].empty()) return 1;
        p[2].push_back(c[0].back());
        c[0].pop_back();
      } else {
        if (c[0].size() >= c[1].size()) {
          // if (c[0].empty()) return 1;
          p[0].push_back(c[0].back());
          c[0].pop_back();
        } else {
          // if (c[1].empty()) return 1;
          p[0].push_back(c[1].back());
          c[1].pop_back();
        }
      }
    }
  }

  vector<int> r(n);
  for (int i = 0; i < n; i++) r[p[(i + 1) % 3][i / 3]] = i + 1;

  for (auto e : r) cout << e << ' ';
  cout << endl;
}