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>
void err() { std::cout << std::endl; }
template <typename T, typename... Args>
void err(T a, Args... args) {
  std::cout << a << ' ';
  err(args...);
}
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 22;
int n, cnt;
int a[maxn];
int tot = 0;
bool isprime(long long a) {
  if (a <= 3) return a > 1;
  if (a % 6 != 1 && a % 6 != 5) return false;
  long long sqr = (long long)sqrt(a);
  for (long long i = 5; i <= sqr; i += 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
long long p = 1;
int main() {
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    if (isprime(i)) {
      for (int j = i; j <= n; j += i) {
        if (a[j] == 0) {
          a[j] = p;
        }
      }
      p++;
    }
  }
  for (long long i = 2; i <= n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}