Problem:
Score : 600 points

Problem Statement

Takahashi has N cards. The i-th of these cards has an integer A_i written on it.

Takahashi will choose an integer K, and then repeat the following operation some number of times:

Choose exactly K cards such that the integers written on them are all different, and eat those cards. (The eaten cards disappear.)

For each K = 1,2, \ldots, N, find the maximum number of times Takahashi can do the operation.

Constraints

1 \le N \le 3 \times 10^5

1 \le A_i \le N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print N integers.
The t-th (1 \le t \le N) of them should be the answer for the case K=t.

Sample Input 1

3
2 1 2

Sample Output 1

3
1
0

For K = 1, we can do the operation as follows:

Choose the first card to eat.

Choose the second card to eat.

Choose the third card to eat.

For K = 2, we can do the operation as follows:

Choose the first and second cards to eat.

For K = 3, we cannot do the operation at all. Note that we cannot choose the first and third cards at the same time.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

5
2
1
1
1

Sample Input 3

4
1 3 3 3

Sample Output 3

4
1
0
0

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.util.stream.Collectors;

class Main {
	public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int N = Integer.parseInt(sc.next());
      List<Integer> list = new ArrayList<>();
      
      for (int i = 0; i < N; ++i) {
        list.add(Integer.parseInt(sc.next()));
      }
      
      Map<Integer, List<Integer>> tempMap = list.stream().collect(Collectors.groupingBy(Integer::intValue));
      List<Integer> tempList = tempMap.values().stream().map(List::size).collect(Collectors.toList());
      tempList.sort(Comparator.comparingInt(Integer::intValue).reversed());
      Integer[] arr = tempList.toArray(new Integer[0]);
      
      for (int i = 1; i <= N; ++i) {
        Integer[] temparr = arr.clone();
        // 最終的な回数
        int result = 0;
        int counter = 0;
        boolean isFinished = false;
        // 明らかに不可能な場合はループを回さない
        if (temparr.length < i) isFinished = true;
        else if (temparr.length == i) {
          result = temparr[i - 1];
          isFinished = true;
        }
        
        // 食べられる間ループする
        while (!isFinished) {
          isFinished = true;
          // 先頭から辿っていって数の多いものからたべていく
          for (int j = 0; j < temparr.length; ++j) {
            if (temparr[j] > 0) {
              // 食べられる場合は1減らしてカウンタをインクリメント
              temparr[j]--;
              counter++;
            }
            if (counter == i) {
              // カウンタがKに達した場合は食べられた回数をインクリメントし、カウンタを初期化して次のループへ
              result++;
              counter = 0;
              isFinished = false;
              break;
            }
          }
          
        }
        
        System.out.println(result);
      }
    }
}