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>
#pragma GCC optimize("O3")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
bool isPrime[100005];
void sie(long long n) {
  memset(isPrime, true, sizeof isPrime);
  isPrime[0] = isPrime[1] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (isPrime[i]) {
      for (long long j = i * i; j <= n; j += i) {
        isPrime[j] = false;
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> ans(n + 1, 0);
  sie(n + 1);
  long long k = 1;
  for (long long i = 2; i <= n; i++) {
    if (isPrime[i]) {
      ans[i] = k;
      for (long long j = i * i; j <= n; j += i) {
        ans[j] = k;
      }
      k++;
    }
  }
  for (long long i = 2; i <= n; i++) cout << ans[i] << " ";
  return 0;
}