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 MAX = 1e9;
const int MOD = 1e9 + 7;
const long long INF = 1e18;
const long double PI = 3.14159265358979323846;
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("no-stack-protector")
template <typename T>
void read(vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) cin >> a[i];
}
template <typename T>
void print(vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) cout << a[i] << " ";
  cout << endl;
}
template <typename T>
void print2(vector<vector<T> >& a) {
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < a[i].size(); ++j) {
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
}
template <typename T>
void read2(vector<vector<T> >& a) {
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < a[i].size(); ++j) {
      cin >> a[i][j];
    }
  }
}
vector<int> resheto(int n) {
  vector<int> primes(n + 1, 0);
  int iter = 1;
  for (int i = 2; i <= n; ++i) {
    if (primes[i] == 0) {
      for (int j = i; j <= n; j += i) {
        primes[j] = iter;
      }
      iter++;
    }
  }
  return primes;
}
void write() {
  int n;
  cin >> n;
  vector<int> a = resheto(n);
  for (int i = 2; i < n + 1; ++i) {
    cout << a[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  for (int T = 0; T < t; ++T) write();
  return 0;
}