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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1;
  while (TESTS--) {
    long long n;
    cin >> n;
    long long minPrime[n + 1];
    for (long long i = 0; i < n + 1; i++) minPrime[i] = 0;
    for (long long i = 2; i * i <= n; ++i) {
      if (minPrime[i] == 0) {
        for (long long j = i * i; j <= n; j += i) {
          if (minPrime[j] == 0) {
            minPrime[j] = i;
          }
        }
      }
    }
    for (long long i = 2; i <= n; ++i) {
      if (minPrime[i] == 0) {
        minPrime[i] = i;
      }
    }
    long long x = 1;
    map<long long, long long> m;
    for (long long i = 2; i < n + 1; i++) {
      if (minPrime[i] == i) {
        cout << x << " ";
        m[i] = x;
        x++;
      } else
        cout << m[minPrime[i]] << " ";
    }
  }
  return 0;
}