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;
const long long int MAX = 1000006;
long long int prime[MAX];
vector<long long int> v;
void Sieve() {
  long long int i, j;
  prime[0] = prime[1] = 1;
  v.push_back(2);
  for (i = 4; i < MAX; i += 2) prime[i] = 1;
  for (i = 3; i * i <= MAX; i += 2) {
    if (prime[i] == 0) {
      for (j = i * i; j <= MAX; j += 2 * i) prime[j] = 1;
    }
  }
  for (i = 3; i <= MAX; i += 2)
    if (prime[i] == 0) v.push_back(i);
}
long long int arr[MAX];
int main() {
  Sieve();
  long long int n;
  scanf("%lld", &(n));
  long long int i, j;
  for (i = 2; i <= n; i += 2) arr[i] = 1;
  for (i = 1; i < v.size(); i++) {
    arr[v[i]] = i + 1;
    for (j = v[i] * v[i]; j <= n; j += (2 * v[i])) arr[j] = i + 1;
  }
  for (i = 2; i <= n; i++) printf("%lld ", arr[i]);
}