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;
using vi = vector<long long int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vld = vector<long double>;
using pii = pair<long long int, long long int>;
using psi = pair<string, long long int>;
using pci = pair<char, long long int>;
using vpii = vector<pii>;
long long int mod = 1e9 + 7;
long long int const maxn = 1e5 + 10;
long long int const inf = 1e18;
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int powm(long long int x, long long int n, long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long int modinverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool prime(long long int x) {
  if (x <= 1) return false;
  for (int i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
long long int divisor(long long int x) {
  long long int cnt = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i != x / i)
        cnt += 2;
      else
        cnt += 1;
    }
  }
  return cnt;
}
vector<long long int> sieve(long long int n) {
  bool prim[n + 1];
  memset(prim, true, sizeof(prim));
  for (long long int p = 2; p * p <= n; p++) {
    if (prim[p] == true) {
      for (int i = p * p; i <= n; i += p) prim[i] = false;
    }
  }
  vector<long long int> v;
  for (int i = 2; i <= n; i++)
    if (prim[i]) v.push_back(i);
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    vi a(maxn, 0);
    long long int x = 0;
    for (int i = 2; i <= n; i++) {
      if (a[i] != 0) continue;
      x++;
      for (int j = i; j <= n; j += i) {
        if (a[j] == 0) a[j] = x;
      }
    }
    for (long long int i = 2; i <= n; i++) cout << a[i] << ' ';
  }
  return 0;
}