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;
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  int a[n + 1];
  int x = 1;
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      a[p] = x;
      prime[p] = false;
      for (int i = p * p; i <= n; i += p) {
        prime[i] = false;
        a[i] = x;
      }
      ++x;
    }
  }
  for (int p = 2; p <= n; p++) {
    if (prime[p])
      cout << x++ << " ";
    else
      cout << a[p] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
  int n;
  cin >> n;
  SieveOfEratosthenes(n);
}