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;
long long int Lpow(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a);
    a = (a * a);
    b = b / 2;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> arr(n + 1, 0);
  long long int curr = 0;
  for (long long int i = 2; i <= n; i++) {
    if (arr[i] == 0) {
      curr++;
      arr[i] = curr;
      for (long long int j = i * i; j <= n; j += i) {
        if (arr[j] == 0) arr[j] = curr;
      }
    }
  }
  for (long long int i = 2; i <= n; i++) cout << arr[i] << " ";
  cout << '\n';
  return 0;
}