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 namespace std;
long long T, t;
void solve() {
  int i, n, flag = 0, a, sum2 = 0, sum = 0, cnt = 0, cnt1 = 0;
  cin >> n;
  bool mark[100002];
  int mark2[100002];
  memset(mark, true, sizeof(mark));
  memset(mark2, -1, sizeof(mark2));
  cnt = 0;
  for (int p = 2; p * p < 100002; p++) {
    if (mark[p] == true) {
      cnt++;
      for (int i = p * 2; i < 100002; i += p) {
        mark[i] = false;
        if (mark2[i] == -1) mark2[i] = cnt;
      }
    }
  }
  cnt = 0;
  for (long long i = 2; i < n + 1; i++) {
    if (mark[i]) {
      cnt++;
      cout << cnt << " ";
    } else
      cout << mark2[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) solve();
  return 0;
}