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: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.NoSuchElementException;

public class Main implements Runnable {

	public static void main(String[] args) {

		new Thread(null, new Main(), "", 16 * 1024 * 1024).start();

	}

	@Override
	public void run() {
		exec();
	}

	private void exec() {

		FastScanner sc = new FastScanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		for(int i = 0 ; i < n ; i++) {
			a[i] = sc.nextInt();
		}
		Arrays.sort(a);

		int[] ans = new int[n];
		ans[0] = n;
		for(int i = 2 ; i <= n ; i++) {
			int[] tmp = Arrays.copyOf(a, a.length);
			int cnt = 0;
			int pre = 0;
			for(int j = 0 ; j < n ; j++) {
				if(tmp[j] != pre) {
					cnt++;
					pre = tmp[j];
					tmp[j] = 0;
				}
				if(cnt == i) {
					ans[i - 1]++;
					cnt = 0;
				}
			}
		}

		for(int i = 0 ; i < n ; i++) {
			System.out.println(ans[i]);
		}

	}

}

class FastScanner {

	private final InputStream in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	public FastScanner(InputStream in) {
		this.in = in;
	}

	private boolean hasNextByte() {
		if(ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if(buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if(hasNextByte()) {
			return buffer[ptr++];
		} else {
			return -1;
		}
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	public boolean hasNext() {
		while(hasNextByte() && !isPrintableChar(buffer[ptr])) {
			ptr++;
		}
		return hasNextByte();
	}

	public String next() {
		if (!hasNext()) throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while(isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			}else if(b == -1 || !isPrintableChar(b)){
				return minus ? -n : n;
			}else{
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
		return (int) nl;
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

}