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;
bool prime(long long int a) {
  if (a == 2)
    return true;
  else if (a == 3)
    return true;
  else {
    int k = 0;
    for (int i = 2; i * i <= a; i++) {
      if (a % i == 0) {
        k = 1;
        break;
      }
    }
    if (k == 0)
      return true;
    else
      return false;
  }
}
long long int factor(long long int a) {
  int k = 0;
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      k = max(k, i);
      if (a / i > k) k = a / i;
    }
  }
  return k;
}
int main() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  a[0] = 1;
  long long int k = 0;
  for (int i = 2; i <= n; i++) {
    if (prime(i)) {
      k++;
      a[i] = k;
    } else
      a[i] = a[factor(i)];
  }
  for (int i = 2; i <= n; i++) cout << a[i] << " ";
}