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 long long mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long b, long long c) {
  long long v = 1;
  for (; b; b >>= 1, a = a * a % c)
    if (b & 1) v = v * a % c;
  return v;
}
long long inv(long long x) { return qpow(x, mod - 2, mod); }
const int maxn = 5 + 1e5;
int n;
vector<int> prime;
int ans[maxn];
int book[maxn];
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    if (book[i]) continue;
    prime.push_back(i);
    for (int j = i; j <= n / i; j++) book[i * j] = 1;
  }
  for (int i = 0; i < prime.size(); i++) {
    for (int t = prime[i]; t <= n; t += prime[i]) {
      ans[t] = i + 1;
    }
  }
  for (int i = 2; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}