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 X[4] = {-1, 1, 0, 0};
int Y[4] = {0, 0, -1, 1};
int p[100005];
void sieve(long long r) {
  for (long long i = 3; i <= r; i += 2) {
    p[i] = 1;
  }
  for (long long i = 3; i <= r; i += 2) {
    if (p[i] == 1)
      for (long long j = i * i; j <= r; j += i) {
        p[j] = 0;
      }
  }
  p[2] = 1;
  p[1] = p[0] = 0;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    sieve(n);
    long long l = 0;
    vector<int> prime;
    for (int i = 2; i <= n; i++) {
      if (p[i]) {
        l++;
        prime.push_back(i);
      }
    }
    long long ans[n + 2];
    for (auto i : prime) {
      for (long long j = i; j <= n; j = j + i) {
        ans[j] = l;
      }
      l--;
    }
    for (int i = 2; i <= n; i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
}