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 isPrime(int n) {
  float n_sqrt;
  if (n == 2 || n == 3) return 1;
  if (n % 6 != 1 && n % 6 != 5) return 0;
  n_sqrt = floor(sqrt((float)n));
  for (int i = 5; i <= n_sqrt; i += 6) {
    if (n % (i) == 0 | n % (i + 2) == 0) return 0;
  }
  return 1;
}
int mapp[100010];
int main() {
  int cnt = 2;
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    if (i % 2 == 0) {
      mapp[i] = 1;
    } else if (isPrime(i)) {
      mapp[i] = cnt;
      for (int j = 1; j * i <= n; j++) {
        mapp[i * j] = cnt;
      }
      cnt++;
    }
  }
  for (int i = 2; i < n; i++) {
    printf("%d ", mapp[i]);
  }
  printf("%d", mapp[n]);
  return 0;
}