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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        // 整数の入力
        int N = sc.nextInt();
        int[] C = new int[N];
        int[] D = new int[N + 1];
        long[] F = new long[N + 1];
        for (int i = 0; i < N; i++) {
            int A = sc.nextInt() - 1;
            C[A]++;
        }
        for (int i = 0; i < N; i++) {
            D[C[i]]++;
        }
        int nowSum = N-D[0];
        long lSum = 0*D[0];
        F[1] = nowSum;
        for (int i = 2; i <= N; i++) {
            nowSum -= D[i-1];
            lSum += (long)(i-1)*(long)D[i-1];
            F[i] = lSum/i + (long)nowSum ;
        }
        int x = N;
        for(int nowK=1;nowK<=N;nowK++) {
            while (x > 0 && F[x] < nowK) {
                x--;
            }
            System.out.println(x);
        }
    }
}