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 arr[100000];
int main() {
  int n, k = 0, i, j, a;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    a = 0;
    for (j = 2; j < 400; j++) {
      if (i % j != 0 || i == j) a++;
    }
    if (a == 398) {
      arr[k] = i;
      k++;
    }
  }
  for (i = 2; i <= n; i++) {
    for (j = 0; j < k; j++) {
      if (i % arr[j] == 0) break;
    }
    printf("%d ", j + 1);
  }
  printf("\n");
}