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>
void read(T &x) {
  cin >> x;
}
template <typename T, typename T0>
void read(T &x, T0 &y) {
  cin >> x >> y;
}
template <typename T, typename T0, typename T1>
void read(T &x, T0 &y, T1 &z) {
  cin >> x >> y >> z;
}
template <typename T, typename T0, typename T1, typename T2>
void read(T &x, T0 &y, T1 &z, T2 &w) {
  cin >> x >> y >> z >> w;
}
template <typename T, typename T0>
void read(pair<T, T0> &p) {
  cin >> p.fst >> p.scd;
}
template <typename T>
void read(vector<T> &oneD) {
  for (long long int i = 0; i < (int)(oneD).size(); i++) {
    read(oneD[i]);
  }
}
template <typename T>
void read(T oneD[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    read(oneD[i]);
  }
}
template <typename T>
void write(T &x) {
  cout << x << " ";
}
template <typename T, typename T0>
void write(T &x, T0 &y) {
  cout << x << " " << y << "\n";
}
template <typename T, typename T0, typename T1>
void write(T &x, T0 &y, T1 &z) {
  cout << x << " " << y << " " << z << "\n";
}
template <typename T, typename T0, typename T1, typename T2>
void write(T &x, T0 &y, T1 &z, T2 &w) {
  cout << x << " " << y << " " << z << " " << w << "\n";
}
template <typename T, typename T0>
void write(pair<T, T0> &p) {
  write(p.fst);
  write(p.scd);
  cout << endl;
}
template <typename T>
void write(vector<T> &oneD) {
  for (long long int i = 0; i < (int)(oneD).size(); i++) {
    write(oneD[i]);
  }
  cout << endl;
}
template <typename T>
void write(T oneD[], int n) {
  for (long long int i = 0; i < n; i++) {
    write(oneD[i]);
  }
  cout << endl;
}
template <typename T, typename T0>
void write(map<T, T0> &mpp) {
  for (auto it : mpp) {
    write(it.fst);
    cout << ": ";
    write(it.scd);
    cout << "\n";
  }
  cout << endl;
}
template <typename T>
inline bool umax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool umin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
template <class T>
void offset(long long int o, T &x) {
  x += o;
}
template <class T>
void offset(long long int o, vector<T> &x) {
  for (auto &a : x) offset(o, a);
}
template <class T, size_t S>
void offset(long long int o, array<T, S> &x) {
  for (auto &a : x) offset(o, a);
}
inline bool isSquare(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool isPowerOfTwo(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
vector<long long int> prime;
bool isPrime[100001];
void sieve(long long int N) {
  for (long long int i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long int i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
  for (long long int i = 2; i <= N; i++) {
    if (isPrime[i] == true) {
      prime.push_back(i);
    }
  }
}
long long int minPrime[(int)(1 * 1e6 + 10)];
void factorSieve(long long int n) {
  memset(minPrime, 0, sizeof(minPrime));
  for (long long int i = 2; i * i <= n; ++i) {
    if (minPrime[i] == 0) {
      for (long long int j = i * i; j <= n; j += i) {
        if (minPrime[j] == 0) {
          minPrime[j] = i;
        }
      }
    }
  }
  for (long long int i = 2; i <= n; ++i) {
    if (minPrime[i] == 0) {
      minPrime[i] = i;
    }
  }
}
long long int factorize(long long int n) {
  map<long long int, long long int> m;
  m.clear();
  while (n != 1) {
    m[minPrime[n]]++;
    n /= minPrime[n];
  }
  for (auto i : m) {
    if (i.second > 1) {
      return 0;
    }
  }
  return 1;
}
string to_string(char c) { return string(1, c); }
const long long int N = 1e8;
void solve() {
  sieve(1e5);
  long long int n, ma = 0;
  read(n);
  long long int ans[n + 1];
  for (long long int i = (2); ((1) > 0 ? i < (n + 1) : i > (n + 1)); i += (1)) {
    if (isPrime[i]) {
      ma++;
      ans[i] = ma;
    } else {
      for (long long int j = 2; j * j <= i; j++) {
        if (i % j == 0) {
          ans[i] = ans[j];
          break;
        }
      }
    }
  }
  for (long long int i = (2); ((1) > 0 ? i < (n + 1) : i > (n + 1)); i += (1)) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  for (long long int i = (0); ((1) > 0 ? i < (t) : i > (t)); i += (1)) {
    solve();
  }
}