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;
int main() {
  int n;
  cin >> n;
  int arr[n + 1];
  memset(arr, 0, sizeof(arr));
  int ans = 1;
  for (int i = 4; i <= n; i = i + 2) arr[i] = 1;
  for (int i = 3; i * i <= n; i += 2) {
    if (arr[i] == 0) {
      for (int j = i * i; j <= n; j += i) {
        arr[j] = 1;
      }
    }
  }
  int dp[n + 1];
  dp[2] = 1;
  dp[3] = 2;
  int k = 3;
  for (int i = 4; i <= n; i++) {
    if (arr[i] == 0) {
      dp[i] = k;
      k++;
    } else if (i % 2 == 0)
      dp[i] = 1;
    else {
      for (int j = 3; j < i; j += 2) {
        if (i % j == 0) {
          dp[i] = dp[j];
          break;
        }
      }
    }
  }
  for (int i = 2; i <= n; i++) cout << dp[i] << ' ';
  return 0;
}