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.

Examples

Input

3
2 1 2


Output

3
1
0


Input

5
1 2 3 4 5


Output

5
2
1
1
1


Input

4
1 3 3 3


Output

4
1
0
0

public class Main {

  private static void solve() {
    int n = ni();

    int[] c = new int[n];
    long[] d = new long[n + 1];
    for (int v : na(n)) {
      v--;
      c[v]++;
      d[c[v]]++;
    }
    long[] sum = new long[n + 1];
    for (int i = 1; i <= n; i++) {
      sum[i] = sum[i - 1] + d[i];
    }

    for (int k = 1; k <= n; k++) {
      int left = 0;
      int right = n + 1;
      while (right - left > 1) {
        int x = (left + right) / 2;
        if (sum[x] >= (long) k * x) {
          left = x;
        } else {
          right = x;
        }
      }
      System.out.println(left);
    }


  }

  public static void main(String[] args) {
    new Thread(null, new Runnable() {
      @Override
      public void run() {
        long start = System.currentTimeMillis();
        String debug = args.length > 0 ? args[0] : null;
        if (debug != null) {
          try {
            is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
          } catch (Exception e) {
            throw new RuntimeException(e);
          }
        }
        reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
        solve();
        out.flush();
        tr((System.currentTimeMillis() - start) + "ms");
      }
    }, "", 64000000).start();
  }

  private static java.io.InputStream is = System.in;
  private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
  private static java.util.StringTokenizer tokenizer = null;
  private static java.io.BufferedReader reader;

  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }

  private static double nd() {
    return Double.parseDouble(next());
  }

  private static long nl() {
    return Long.parseLong(next());
  }

  private static int[] na(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = ni();
    return a;
  }

  private static char[] ns() {
    return next().toCharArray();
  }

  private static long[] nal(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nl();
    return a;
  }

  private static int[][] ntable(int n, int m) {
    int[][] table = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[i][j] = ni();
      }
    }
    return table;
  }

  private static int[][] nlist(int n, int m) {
    int[][] table = new int[m][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[j][i] = ni();
      }
    }
    return table;
  }

  private static int ni() {
    return Integer.parseInt(next());
  }

  private static void tr(Object... o) {
    if (is != System.in)
      System.out.println(java.util.Arrays.deepToString(o));
  }
}

