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;
const int N = 1e5 + 5;
const long long int infP = 1e18;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> div(n + 1);
  for (long long int i = 2; i <= n; i++) {
    if (div[i] == 0) {
      for (long long int j = i; j <= n; j += i) div[j] = i;
    }
  }
  long long int val = 1;
  vector<long long int> v(n + 1);
  for (long long int i = 2; i <= n; i++) {
    if (div[i] == i) v[i] = val++;
  }
  for (long long int i = 2; i <= n; i++) cout << v[div[i]] << " ";
}
void debug(long long int tt) {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}