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 int N = 1e5 + 9, INF = 0x3f3f3f3f;
int s[N];
inline long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    s *= 10;
    s += ch - '0';
    ch = getchar();
  }
  return s * f;
}
bool qwq(int n) {
  int x = sqrt(n);
  for (int i = 2; i <= x; ++i) {
    if (n % i == 0) return false;
  }
  return true;
}
int main(void) {
  int n = read();
  s[2] = 1;
  int num = 1;
  for (int i = 3; i <= n; ++i) {
    if (qwq(i)) {
      ++num;
      s[i] = num;
    } else {
      for (int j = 2; j <= n; ++j) {
        if (i % j == 0) {
          s[i] = s[j];
          break;
        }
      }
    }
  }
  for (int i = 2; i <= n; ++i) printf("%d ", s[i]);
  return 0;
}