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 int maxn = 100005;
int isprime[maxn];
int prime[maxn];
int ans[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= maxn; i++) isprime[i] = 1;
  for (int i = 2; i * i <= maxn; i++) {
    if (isprime[i]) {
      for (int j = i * i; j <= maxn; j += i) isprime[j] = 0;
    }
  }
  int cnt = 0;
  for (int i = 0; i <= maxn; i++)
    if (isprime[i]) {
      prime[cnt++] = i;
    }
  int num = 0;
  for (int i = 2; i <= n; i++) {
    if (isprime[i])
      ans[i] = ++num;
    else {
      for (int j = 0; j <= cnt; j++)
        if (i % prime[j] == 0) {
          ans[i] = j + 1;
          break;
        }
    }
  }
  for (int i = 2; i <= n; i++) {
    cout << ans[i];
    if (i < n) cout << ' ';
  }
  cout << endl;
  return 0;
}