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;
bitset<100010> vis;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long mcm(long long a, long long b) { return (b * a) / gcd(a, b); }
bool comp(int a, int b) { return a > b; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  int cont = 1;
  vis.set();
  int arr[n + 10];
  arr[0] = arr[1] = 0;
  for (long long i = 2; i < n + 5; ++i) {
    if (vis[i]) {
      arr[i] = cont;
      for (long long j = i * i; j < n + 5; j += i) {
        vis[j] = 0;
        arr[j] = cont;
      }
      cont++;
    }
  }
  for (int i = 2; i <= n; ++i) {
    cout << arr[i] << " ";
  }
  cout << "\n";
  return 0;
}