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>
const int maxn = 1e5 + 5;
int main() {
  int n;
  scanf("%d", &n);
  bool isP[maxn];
  int P[maxn], ans[maxn];
  int total = 0;
  memset(isP, true, sizeof(isP));
  memset(P, 0, sizeof(P));
  memset(ans, 0, sizeof(ans));
  isP[0] = isP[1] = false;
  for (int i = 2; i <= n; i++) {
    if (isP[i]) {
      P[++total] = i;
      ans[i] = total;
    }
    for (int j = 1; j <= total && i * P[j] <= n; j++) {
      isP[i * P[j]] = false;
      ans[i * P[j]] = ans[P[j]];
      if (!(i % P[j])) break;
    }
  }
  for (int i = 2; i <= n; i++) {
    if (i != 2) putchar(' ');
    printf("%d", ans[i]);
  }
  return 0;
}