Problem: You're given an integer n. For every integer i from 2 to n, assign a positive integer a_i such that the following conditions hold:

  * For any pair of integers (i,j), if i and j are coprime, a_i ≠ a_j. 
  * The maximal value of all a_i should be minimized (that is, as small as possible). 



A pair of integers is called [coprime](https://en.wikipedia.org/wiki/Coprime_integers) if their [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) is 1.

Input

The only line contains the integer n (2 ≤ n ≤ 10^5).

Output

Print n-1 integers, a_2, a_3, …, a_n (1 ≤ a_i ≤ n). 

If there are multiple solutions, print any of them.

Examples

Input


4


Output


1 2 1 

Input


3


Output


2 1

Note

In the first example, notice that 3 and 4 are coprime, so a_3 ≠ a_4. Also, notice that a=[1,2,3] satisfies the first condition, but it's not a correct answer because its maximal value is 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  long a[100001] = {0};
  long b[100001] = {0};
  int shu = 0;
  long c[100001] = {0};
  int p = 0;
  int i, j;
  int n;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    if (a[i] == 0) {
      j = i;
      while (j <= n) {
        if (a[i] == 0) a[j + i] = 1;
        j += i;
      }
    }
  }
  for (i = 2; i <= n; i++)
    if (a[i] == 0) {
      b[shu] = i;
      shu++;
    }
  int l;
  for (l = 0; l < shu; l++) {
    p++;
    for (i = b[l]; i <= n; i += b[l]) {
      if (c[i] == 0) c[i] = p;
    }
  }
  for (i = 2; i <= n; i++) {
    printf("%ld", c[i]);
    if (i < n) printf(" ");
  }
  printf("\n");
}