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;
vector<int> build(int n) {
  map<int, bool> s;
  for (int i = 2; i * i <= n; i++) {
    if (s[i] == false) {
      for (int j = 2 * i; j <= n; j = j + i) s[j] = true;
    }
  }
  vector<int> t;
  for (int i = 2; i <= n; i++) {
    if (s[i] == false) t.push_back(i);
  }
  return t;
}
int main() {
  int n;
  cin >> n;
  vector<int> a = build(n);
  int b[n - 1];
  int s = 1;
  for (int i = 0; i < a.size(); i++) {
    for (int j = a[i]; j <= n; j = j + a[i]) b[j - 2] = s;
    s++;
  }
  for (int i = 0; i < n - 1; i++) cout << b[i] << " ";
  return 0;
}