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;
inline int read() {
  int ok = 1, k = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') ok = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    k = k * 10 + c - '0';
    c = getchar();
  }
  return ok * k;
}
int s[100005];
bool tot[100005];
int x = 0;
void work(int n) {
  for (int i = 2; i <= n; i++) {
    if (tot[i]) continue;
    for (int j = 2; j * j <= i; j++)
      if (i % j == 0) tot[i] = 1;
    if (!tot[i]) x++;
    s[i] = x;
    for (int j = 2; j * i <= n; j++)
      tot[j * i] = 1, s[j * i] = min(s[j * i], x);
  }
}
int main() {
  int n = read();
  memset(s, 0x3f, sizeof(s));
  work(n);
  for (int i = 2; i <= n; i++) {
    printf("%d ", s[i]);
  }
  return 0;
}