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;
bool cmp(pair<int, int> p1, pair<int, int> p2) {
  if (p1.first > p2.first) {
    return p1.first;
  } else {
    return p2.first;
  }
}
bool comp(pair<int, int> p1, pair<int, int> p2) {
  if (p1.first == p2.first) {
    if (p1.second < p2.second) {
      return true;
    } else {
      return false;
    }
  } else {
    return false;
  }
}
int arr[100009];
void seive() {
  memset(arr, 0, sizeof(arr));
  for (int i = 2; i <= 1000; i++) {
    if (arr[i] == 0) {
      for (int j = i * i; j <= 100000; j += i) {
        arr[j] = 1;
      }
    }
  }
  arr[1] = 1;
}
int main() {
  int n;
  cin >> n;
  seive();
  int a[100009];
  int odd = 2;
  for (int i = 2; i <= n; i++) {
    if (i % 2 == 0) {
      cout << 1 << " ";
    } else if (arr[i] == 0) {
      a[i] = odd;
      cout << odd++ << " ";
    } else {
      for (int j = 2; j <= sqrt(i); j++) {
        if (i % j == 0) {
          cout << a[j] << " ";
          break;
        }
      }
    }
  }
}