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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val list = readLine()!!.split(" ").map { it.toInt() }
    val m = HashMap<Int, Int>()
    list.forEach {
        if (m.containsKey(it)) {
            m[it] = m[it]!! + 1
        } else {
            m[it] = 1
        }
    }
    val s = m.map { it.value }.sortedDescending()

    (1..n).forEach {
        println(count(s, it))
    }
}

fun count(list: List<Int>, x: Int): Int {
    val q = PriorityQueue<Int>(list.size, Comparator<Int> { a, b -> b.compareTo(a) })
    q.addAll(list)

    var r = 0
    while (q.size >= x) {
        val k = ArrayList<Int>()
        repeat(x) {
            k.add(q.poll())
        }
        val last = k.last()
        r += last

        k.map { it - last }.filter { it > 0 }.forEach {
            q.add(it)
        }
    }
    return r
}