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;
unordered_map<long long, long long> m;
const long long N = 1e5 + 10;
int n, t;
long long a[100010];
int prime[N], check[N], sum[N], cnt;
inline void init(int n) {
  int i, j;
  int cut = 0;
  for (i = 2; i <= n; i++) {
    if (check[i] == 0) prime[++cnt] = i, sum[i] = cnt;
    for (j = 1; j <= cnt && prime[j] * i <= n + 1; j++) {
      check[prime[j] * i] = 1;
      sum[prime[j] * i] = min(sum[i], sum[prime[j]]);
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n; i++) {
    sum[i] = N + 10;
    if (i % 2 == 0) {
      sum[i] = 1;
    }
  }
  init(n);
  for (int i = 2; i < n; i++) {
    cout << sum[i] << ' ';
  }
  cout << sum[n] << endl;
  return 0;
}