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;
const int MAXN = 2e6;
int n, res[MAXN];
vector<bool> isprime(MAXN, true);
vector<int> prime, SPF(MAXN);
void criba(int n) {
  int i, j;
  isprime[0] = isprime[1] = false;
  for (i = 2; i <= n; i++) {
    if (isprime[i]) {
      prime.push_back(i);
      SPF[i] = i;
    }
    for (j = 0; j < prime.size() && i * prime[j] <= n && prime[j] <= SPF[i];
         j++) {
      isprime[i * prime[j]] = false;
      SPF[i * prime[j]] = prime[j];
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  criba(2000000);
  cin >> n;
  res[2] = 1;
  int aux = 2;
  for (int i = 1; i < min((int)prime.size(), n + 1); i++) {
    res[prime[i]] = aux;
    aux++;
    int k = 2;
    while (k * prime[i] <= n) {
      res[k * prime[i]] = res[prime[i]];
      k++;
    }
  }
  int k = 2;
  while (k <= n) {
    res[k] = 1;
    k += 2;
  }
  for (int i = 2; i < n + 1; i++) cout << res[i] << " ";
  cout << endl;
}