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;
template <typename T>
void ri(T& x);
const int MAXN = 1e5 + 2;
int n, a[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int cur = 1;
  for (int i = 2; i <= n; ++i) {
    if (a[i] == 0) {
      a[i] = cur;
      for (int j = i << 1; j <= n; j += i) {
        a[j] = cur;
      }
      ++cur;
    }
  }
  for (int i = 2; i <= n; ++i) {
    cout << a[i] << ' ';
  }
  return 0;
}
template <typename T>
void ri(T& x) {
  char c;
  bool minus = false;
  while ((c = getchar()) < '0' || '9' < c)
    if (c == '-') minus = true;
  x = c - '0';
  while ('0' <= (c = getchar()) && c <= '9') x = 10 * x - '0' + c;
  if (minus) x = -x;
}