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 MAXN = 1e6;
int n;
bool isPrime[MAXN + 10];
int small[MAXN + 10];
vector<int> primeList;
int cnt[MAXN + 10];
void sieve() {
  int i, j;
  for (i = 2; i <= MAXN; i++) isPrime[i] = true;
  for (i = 2; i * i <= MAXN; i++) {
    if (!isPrime[i]) continue;
    small[i] = i;
    for (j = i * i; j <= MAXN; j += i) isPrime[j] = false, small[j] = i;
  }
  int k = 1;
  for (i = 2; i <= MAXN; i++)
    if (isPrime[i]) primeList.push_back(i), cnt[i] = k++;
  for (i = 2; i <= MAXN; i++)
    if (small[i] == 0) small[i] = i;
}
int main() {
  int i, j;
  sieve();
  scanf("%d", &n);
  for (i = 2; i <= n; i++) printf("%d ", cnt[small[i]]);
}