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>
long long fastpow(long long a, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    int k = fastpow(a, n / 2);
    return k * k;
  } else
    return a * fastpow(a, n - 1);
}
int abs(int a) {
  if (a < 0) return a * (-1);
  return a;
}
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
int main() {
  int n;
  std::cin >> n;
  int a[1000000];
  for (int i = 0; i < 1000000; ++i) a[i] = 0;
  int k = 1;
  for (int i = 2; i <= n; ++i)
    if (a[i] == 0) {
      for (int j = i; j <= n; j += i) {
        a[j] = k;
      }
      k++;
    }
  for (int i = 2; i <= n; ++i) std::cout << a[i] << " ";
  return 0;
}