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>
const int INF = 1e8;
using namespace std;
int criba[2000002];
void pre() {
  criba[1] = 1;
  for (long long int i = 4; i <= 200000; i += 2) criba[i] = 1;
  for (long long int i = 3; i <= 200000; i += 2) {
    if (criba[i] == 0) {
      for (long long int j = i * i; j <= 200000; j += (2 * i)) {
        criba[j] = 1;
      }
    }
  }
}
bool isPrime(int i) {
  if (criba[i] == 0) return true;
  return false;
}
int ans[200000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  vector<int> primes;
  cin >> n;
  pre();
  for (int i = 1; i < n + 1; i++) ans[i] = -1;
  int x = 1;
  ans[1] = 1;
  for (int i = 2; i < n + 1; i++) {
    if (isPrime(i)) {
      ans[i] = x;
      x++;
      primes.push_back(i);
    }
  }
  for (int i = 0; i < primes.size(); i++) {
    int j = primes[i];
    while (j <= n) {
      if (ans[j] == -1) ans[j] = ans[primes[i]];
      j += primes[i];
    }
  }
  for (int i = 2; i < n + 1; i++) cout << ans[i] << " ";
  return 0;
}