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 main() {
  int n = 0, tw = 1, th = 2, fi = 3, k = 4, p = 0, j;
  int i = 0;
  cin >> n;
  vector<int> a(n + 1, 0);
  for (i = 2; i <= n; i++) {
    if (a[i] != 0)
      continue;
    else if (i % 2 == 0) {
      a[i] = tw;
      j = 2 * i;
      p = 3;
      while (j <= n) {
        a[j] = tw;
        j = i * p;
        p++;
      }
    } else if (i % 3 == 0) {
      a[i] = th;
      j = 2 * i;
      p = 3;
      while (j <= n) {
        a[j] = th;
        j = i * p;
        p++;
      }
    } else if (i % 5 == 0) {
      a[i] = fi;
      j = 2 * i;
      p = 3;
      while (j <= n) {
        a[j] = fi;
        j = i * p;
        p++;
      }
    } else {
      a[i] = k++;
      j = 2 * i;
      p = 3;
      while (j <= n) {
        a[j] = k - 1;
        j = i * p;
        p++;
      }
    }
  }
  for (i = 2; i <= n; i++) cout << a[i] << " ";
  return 0;
}