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;
inline int gi() {
  char c;
  int x = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  return x * f;
}
const int N = 2e5 + 5;
int a[N];
bool p[N];
int main() {
  int n;
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 2; i * i <= n; ++i) {
    if (!p[i]) {
      a[i] = ++cnt;
      for (int j = i * i; j <= n; j += i) p[j] = true, a[j] = cnt;
    }
  }
  for (int i = 2; i <= n; ++i) printf("%d ", a[i] ? a[i] : ++cnt);
}