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;
namespace my_stl {
inline int read() {
  int x = 0, flag = 1;
  char s;
  while (((s = getchar()) < '0' || s > '9') && s ^ '-')
    ;
  if (s == '-') flag = -1, s = getchar();
  do {
    x = (x << 1) + (x << 3) + (s ^ 48);
  } while ((s = getchar()) >= '0' && s <= '9');
  return flag * x;
}
inline long long readl() {
  long long x = 0, flag = 1LL;
  char s;
  while (((s = getchar()) < '0' || s > '9') && s ^ '-')
    ;
  if (s == '-') flag = -1LL, s = getchar();
  do {
    x = (x << 1) + (x << 3) + (s ^ 48);
  } while ((s = getchar()) >= '0' && s <= '9');
  return flag * x;
}
inline void outit(long long num) {
  if (num >= 10) outit(num / 10);
  putchar(num % 10 + '0');
}
inline void P(long long x, char s) {
  x < 0 ? putchar('-'), outit(-x) : outit(x);
  putchar(s);
}
}  // namespace my_stl
using namespace my_stl;
const int N = 1e5 + 10, inf = 2147483645;
int n = read(), a[N], po;
int main() {
  for (register int i = 2; i <= n; ++i) {
    if (!a[i]) {
      a[i] = ++po;
      for (register int j = 2; (j * i) <= n; ++j) {
        a[j * i] = po;
      }
    }
    P(a[i], ' ');
  }
  return 0;
}