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;
using namespace std;
int n;
const long long N = 300005;
long long prime[N] = {0}, num_prime = 0;
int isNotPrime[N] = {1, 1};
void getPrim() {
  for (long long i = 2; i < N; i++) {
    if (!isNotPrime[i]) prime[num_prime++] = i;
    for (long long j = 0; j < num_prime && i * prime[j] < N; j++) {
      isNotPrime[i * prime[j]] = 1;
      if (!(i % prime[j])) break;
    }
  }
}
int num[200005];
int main(int argc, char *argv[]) {
  getPrim();
  while (cin >> n) {
    vector<int> ans;
    int now = 0;
    for (int i = 2; i <= n; i++) {
      if (!isNotPrime[i]) {
        now += 1;
        num[i] = now;
        ans.push_back(now);
      } else {
        for (int j = 2; j * j <= i; j++) {
          if (i % j == 0) {
            ans.push_back(num[j]);
            break;
          }
        }
      }
    }
    for (int i = 0; i < ans.size(); i++) {
      printf("%d", ans[i]);
      if (i != ans.size() - 1) {
        printf(" ");
      }
    }
    cout << endl;
  }
  return 0;
}