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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename P>
P dectobin(P a) {
  if (a == 0)
    return 0;
  else
    return (a % 2 + 10 * dectobin(a / 2));
}
template <typename Y>
Y bintodec(Y a) {
  long long ans = 0, b = 1, t = a;
  while (t) {
    long long ld = t % 10;
    t /= 10;
    ans += ld * b;
    b = b * 2;
  }
  return ans;
}
template <typename H>
H modExp(H x, H n) {
  long long r = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      r = (r * x) % 1000000007;
    }
    x = (x * x) % 1000000007;
    n /= 2;
  }
  return r;
}
template <typename T>
T isPowerOfTwo(T x) {
  return x && (!(x & (x - 1)));
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
void sieve(long long n) {
  long long prime[n + 1];
  long long cnt = 0;
  memset(prime, 0, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == 0) {
      cnt++;
      for (long long i = p * p; i <= n; i += p) prime[i] = cnt;
    }
  }
  cnt = 1;
  for (long long(i) = (2); i < (n + 1); i++) {
    if (prime[i] == 0) {
      prime[i] = cnt;
      cnt++;
    }
  }
  for (long long(i) = (2); i < (n + 1); i++) {
    cout << prime[i] << " ";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ostream_iterator<long long> output(cout, " ");
  long long n;
  cin >> n;
  sieve(n);
  return 0;
}