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 long long int Mod = 998244353;
const long long int mod = 1e9 + 7;
long long int Pow(long long int a, long long int b) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return (res % mod);
}
long long int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
void solve() {
  int n;
  cin >> n;
  int arr[n + 1];
  arr[0] = 0;
  arr[1] = 0;
  memset(arr, 0, sizeof(arr));
  int val = 1;
  if (n == 3)
    cout << 2 << " " << 1;
  else if (n == 2)
    cout << 1;
  else {
    for (int i = 2; i <= sqrt(n); i++) {
      if (arr[i] == 0) {
        arr[i] = val;
        for (int j = i * i; j <= n; j += i) {
          arr[j] = val;
        }
        val++;
      }
    }
    int x = *max_element(arr, arr + (n + 1));
    x++;
    for (int i = sqrt(n); i <= n; i++) {
      if (arr[i] == 0) {
        arr[i] = x;
        x++;
      }
    }
    for (int i = 2; i <= n; i++) {
      cout << arr[i] << " ";
    }
  }
}
int main() {
  solve();
  return 0;
}