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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int arr[n - 1];
  vector<pair<int, int>> primes;
  int next_num = 1;
  for (int i = 2; i <= n; ++i) {
    bool done = false;
    for (auto x : primes) {
      if (i % x.first == 0) {
        arr[i - 2] = x.second;
        done = true;
        break;
      }
    }
    if (!done) {
      arr[i - 2] = next_num;
      primes.push_back({i, next_num});
      next_num++;
    }
  }
  if (n > 0) {
    for (int i = 0; i < n - 2; ++i) {
      cout << arr[i] << " ";
    }
    cout << arr[n - 2] << "\n";
  }
}