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) {
  if (n <= 1) return 1;
  for (int i = 2; i <= ceil(sqrt(n)); i++) {
    if (n % i == 0) return i;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, k = 0, n, m, q = 2;
  cin >> n;
  vector<long long int> v(n + 3, 0);
  v[0] = v[1] = v[2] = 1;
  for (i = 2; i < n - 1 + 2; i++) {
    if (isPrime(i)) {
      v[i] = v[isPrime(i)];
    } else {
      v[i] = q;
      q++;
    }
  }
  for (i = 2; i < n - 1 + 2; i++) cout << v[i] << " ";
  cout << "\n";
  return 0;
}