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;
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
long long T, n, dp[100005];
void solve() {
  cin >> n;
  long long cc = 0;
  for (int i = 2; i <= n; i++) {
    if (dp[i]) continue;
    cc++;
    dp[i] = cc;
    for (int j = i; j <= n; j += i) {
      if (dp[j]) continue;
      dp[j] = cc;
    }
  }
  for (int i = 2; i <= n; i++) cout << dp[i] << ' ';
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  T = 1;
  for (int t = 1; t <= T; t++) solve();
  cout.flush();
  return 0;
}