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;
bool is_prime(int n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  for (int i = 3; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T = 1;
  for (int t = 1; t <= T; t++) {
    int n;
    cin >> n;
    int count = 1;
    int pre[n + 1];
    for (int64_t i = 2; i <= n; i++) {
      if (is_prime(i)) {
        pre[i] = count++;
        for (int64_t j = i * i; j <= n; j += i) {
          pre[j] = pre[i];
        }
      }
    }
    for (int i = 2; i <= n; i++) {
      cout << pre[i] << " ";
    }
  }
}