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;
const int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = (int)1e5 + 10;
const long long int lim = 1e5 + 10;
bool is_prime[MAX + 10];
int resp[MAX + 10];
void crivo() {
  for (int i = 0; i < MAX; i++) {
    is_prime[i] = 1;
  }
  is_prime[0] = is_prime[1] = 0;
  long long int num = 0;
  for (long long int i = 2; i <= lim; i++) {
    if (is_prime[i]) {
      num++;
      resp[i] = num;
      for (long long int j = i * i; j <= lim; j += i) {
        if (is_prime[j] == 1) {
          resp[j] = num;
        }
        is_prime[j] = 0;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  crivo();
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cout << resp[i] << " ";
  }
  cout << endl;
  return 0;
}