Problem: We'll call a set of positive integers a beautiful if the following condition fulfills: for any prime p, if <image>, then <image>. In other words, if one number from the set is divisible by prime p, then at least half of numbers from the set is divisible by p.

Your task is to find any beautiful set, where the number of elements is equal to k and each element doesn't exceed 2k2.

Input

The first line contains integer k (10 ≤ k ≤ 5000) that shows how many numbers the required beautiful set should have.

Output

In the first line print k space-separated integers that are a beautiful set. If there are multiple such sets, you are allowed to print any of them.

Examples

Input

10


Output

16 18 24 27 36 48 54 72 108 144 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O500")
#pragma comment(linker, "/STACK:1677777216")
#pragma warning(default : 4)
using namespace std;
const double eps = 1e-12;
const int oo = 0x3F3F3F3F;
const long long ooLL = 0x3F3F3F3F3F3F3F3FLL;
const int MOD = 1000000007;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void debpr(const T &);
template <typename T1, typename T2>
void debpr(const pair<T1, T2> &);
template <typename T1, typename T2>
void debpr(const vector<T1, T2> &);
template <typename T>
void debpr(const set<T> &);
template <typename T1, typename T2>
void debpr(const map<T1, T2> &);
template <typename T>
void prcont(T be, T en, const string &st, const string &fi, const string &mi) {
  debpr(st);
  bool ft = 0;
  while (be != en) {
    if (ft) debpr(mi);
    ft = 1;
    debpr(*be);
    ++be;
  }
  debpr(fi);
}
template <typename T>
void debpr(const T &a) {}
template <typename T1, typename T2>
void debpr(const pair<T1, T2> &p) {
  debpr("(");
  debpr(p.first);
  debpr(", ");
  debpr(p.second);
  debpr(")");
}
template <typename T1, typename T2>
void debpr(const vector<T1, T2> &a) {
  prcont(a.begin(), a.end(), "[", "]", ", ");
}
template <typename T>
void debpr(const set<T> &a) {
  prcont(a.begin(), a.end(), "{", "}", ", ");
}
template <typename T1, typename T2>
void debpr(const map<T1, T2> &a) {
  prcont(a.begin(), a.end(), "{", "}", ", ");
}
void deb(){};
template <typename T1>
void deb(const T1 &t1) {
  debpr(t1);
  debpr('\n');
}
template <typename T1, typename T2>
void deb(const T1 &t1, const T2 &t2) {
  debpr(t1);
  debpr(' ');
  debpr(t2);
  debpr('\n');
}
template <typename T1, typename T2, typename T3>
void deb(const T1 &t1, const T2 &t2, const T3 &t3) {
  debpr(t1);
  debpr(' ');
  debpr(t2);
  debpr(' ');
  debpr(t3);
  debpr('\n');
}
template <typename T1, typename T2, typename T3, typename T4>
void deb(const T1 &t1, const T2 &t2, const T3 &t3, const T4 &t4) {
  debpr(t1);
  debpr(' ');
  debpr(t2);
  debpr(' ');
  debpr(t3);
  debpr(' ');
  debpr(t4);
  debpr('\n');
}
const double PI = acos(-1.);
long long Round(double x) { return x < 0 ? x - .5 : x + .5; }
template <typename T>
void ass(bool v, const T &x, string m = "Fail") {
  if (!v) {
    deb(m);
    deb(x);
    throw;
  }
}
int main() {
  void run();
  run();
  return 0;
}
const int T = 1 << 15;
int k, p[T], ps, a[T], d[T];
bool isPr(int n) {
  for (int i = 2; i * i <= n; ++i)
    if (n % i == 0) return 0;
  return 1;
}
void run() {
  scanf("%d", &k);
  for (int i = (2), _b(T); i < _b; ++i)
    if (isPr(i)) p[ps++] = i;
  int n = 0, j = 0;
  a[n++] = 1;
  while (n < k) {
    int n0 = n;
    sort(a, a + n0);
    int cp = p[j++], c = cp;
    while (c <= 2 * k * k) {
      int i = 0;
      while (i < n0 && a[i] * c < 2 * k * k) a[n++] = a[i] * c, ++i;
      c *= cp;
    }
  }
  do {
    memset(d, 0, sizeof(d));
    random_shuffle(a, a + n);
    for (int i = (0), _b(k); i < _b; ++i)
      for (int t = (0), _b(j); t < _b; ++t)
        if (a[i] % p[t] == 0) ++d[t];
  } while (*min_element(d, d + j) < (k + 1) / 2);
  deb(n, j);
  for (int i = (0), _b(k); i < _b; ++i) printf("%d%c", a[i], "\n "[i + 1 < k]);
}