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 LLINF = 1e18;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos(-1);
const int SIZE = 200005;
int N;
int A[SIZE];
void solve() {
  int tt = 1;
  for (int i = 0; i < SIZE; i++) {
    A[i] = i;
  }
  for (int i = 2; i * i < SIZE; i++) {
    if (A[i] == i) {
      A[i] = tt;
      for (int j = i + i; j < SIZE; j += i) {
        A[j] = tt;
      }
      tt++;
    }
  }
  for (int i = 2; i < SIZE; i++) {
    if (A[i] == i) A[i] = tt++;
  }
}
int main() {
  solve();
  scanf("%d", &N);
  for (int i = 2; i <= N; i++) {
    printf("%d%c", A[i], i != N ? ' ' : '\n');
  }
  return 0;
}