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 prime(long long n, bool p[], long long ans[]) {
  long long count = 1, i, j;
  for (i = 2; i * i <= n; i++) {
    if (p[i] == true) {
      ans[i] = count;
      for (j = i * i; j <= n; j += i) {
        if (p[j] == true) {
          p[j] = false;
          ans[j] = count;
        }
      }
      count++;
    }
  }
  for (i = 2; i < n; i++) {
    if (ans[i] == 0) {
      ans[i] = count;
      count++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, i;
  cin >> n;
  bool p[n + 1];
  memset(p, true, sizeof(p));
  long long ans[n + 1];
  memset(ans, 0, sizeof(ans));
  prime(n + 1, p, ans);
  for (i = 2; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}