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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
const int MOD = 1e9 + 7;
inline void madd(int& x, int y) {
  x += y;
  if (x >= MOD) {
    x -= MOD;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n;
  cin >> n;
  int count = 1;
  vector<int> ans(n + 1, -1);
  vector<bool> is_prime(n + 1, true);
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i; j <= n; j += i) {
        is_prime[j] = false;
        ans[j] = count;
      }
      count++;
    }
  }
  for (int i = 2; i < n + 1; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}