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;
int main() {
  long long i, j, k, n, t, m, cnt = 0, ans, t2, t1, t3;
  t = 1;
  long long z = 4294967295;
  while (t--) {
    ans = 0;
    cnt = 1;
    t1 = 0;
    t2 = 0;
    t3 = 0;
    cin >> n;
    string s;
    bool p[n + 1];
    long long a[n + 1];
    for (i = 0; i <= n; i++) p[i] = true;
    k = 1;
    for (i = 2; i <= n; i++) {
      if (p[i]) {
        a[i] = k;
        k++;
        for (j = i * i; j <= n; j += i) {
          p[j] = false;
          a[j] = a[i];
        }
      }
    }
    for (i = 2; i <= n; i++) cout << a[i] << " ";
  }
  return 0;
}