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>
using namespace std;
int a[100005] = {0}, b[100005] = {0}, c[100005] = {0};
int main() {
  int i, n, j = 0, k = 1, x, flag;
  scanf("%d", &n);
  b[0] = 2;
  c[0] = x = 1;
  for (i = 2; i <= n; i++) {
    flag = 0;
    for (j = 0; b[j] != 0; j++) {
      if (i % b[j] == 0) {
        a[i] = c[j];
        flag = 1;
        break;
      }
    }
    if (!flag) {
      b[j] = i;
      c[j] = ++x;
      a[i] = c[j];
    }
  }
  printf("%d", a[2]);
  for (i = 3; i <= n; i++) {
    printf(" %d", a[i]);
  }
  printf("\n");
  return 0;
}