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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
vector<long long> divisor(long long n) {
  vector<long long> ret;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}
long long mod = 1e9 + 7;
struct LazySegmentTree {
 private:
  int n;
  vector<int> node, lazy;
  vector<bool> lazyFlag;

 public:
  LazySegmentTree(vector<int> v) {
    int sz = (int)v.size();
    n = 1;
    while (n < sz) n *= 2;
    node.resize(2 * n - 1);
    lazy.resize(2 * n - 1, INT_MAX);
    lazyFlag.resize(2 * n - 1, false);
    for (int i = 0; i < sz; i++) node[i + n - 1] = v[i];
    for (int i = n - 2; i >= 0; i--)
      node[i] = min(node[i * 2 + 1], node[i * 2 + 2]);
  }
  void lazyEvaluate(int k, int l, int r) {
    if (lazyFlag[k]) {
      node[k] = lazy[k];
      if (r - l > 1) {
        lazy[k * 2 + 1] = lazy[k * 2 + 2] = lazy[k];
        lazyFlag[k * 2 + 1] = lazyFlag[k * 2 + 2] = true;
      }
      lazyFlag[k] = false;
    }
  }
  void update(int a, int b, int x, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    lazyEvaluate(k, l, r);
    if (b <= l || r <= a) return;
    if (a <= l && r <= b) {
      lazy[k] = x;
      lazyFlag[k] = true;
      lazyEvaluate(k, l, r);
    } else {
      update(a, b, x, 2 * k + 1, l, (l + r) / 2);
      update(a, b, x, 2 * k + 2, (l + r) / 2, r);
      node[k] = min(node[2 * k + 1], node[2 * k + 2]);
    }
  }
  int find(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    lazyEvaluate(k, l, r);
    if (b <= l || r <= a) return INT_MAX;
    if (a <= l && r <= b) return node[k];
    int vl = find(a, b, 2 * k + 1, l, (l + r) / 2);
    int vr = find(a, b, 2 * k + 2, (l + r) / 2, r);
    return min(vl, vr);
  }
};
int n;
int ans[100004] = {0};
int cnt = 1;
void f() {
  for (int i = 2; i <= n; i++) {
    if (ans[i] == 0) {
      for (int x = i; x <= n; x += i) {
        if (ans[x] == 0) ans[x] = cnt;
      }
      cnt += 1;
    }
  }
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  cin >> n;
  f();
  for (int i = 2; i <= n; i++) {
    cout << ans[i] << " ";
  }
}