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 ll = long long;
using ull = unsigned long long;
using str = string;
using srt = short;
const int INF = (int)1e9 + 7;
const int mod = 998244353;
const ll inf = (ll)1e18;
const ull oo = 1000000000000001;
const int N = 200005;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  int cnt = 1;
  vector<int> pr;
  map<int, bool> p;
  for (int i = 2; i <= n; ++i) {
    bool ok = 1;
    for (int j = 0; ok && j < pr.size() && 1LL * pr[j] * pr[j] <= i; ++j) {
      ok &= (i % pr[j] != 0);
    }
    if (ok) {
      pr.push_back(i);
      p[i] = 1;
    }
  }
  vector<int> ans(n + 1);
  for (int i = 2; i <= n; ++i) {
    if (p[i])
      ans[i] = cnt++;
    else {
      for (int j = 2; j * j <= i; ++j) {
        if (i % j == 0) {
          ans[i] = ans[i / j];
          break;
        }
      }
    }
  }
  for (int i = 2; i <= n; ++i) {
    cout << ans[i] << ' ';
  }
}