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>
using namespace std;
const double Pi = acos(-1.0);
struct no {
  int mask;
  long long num;
  no(int mask = 0, long long num = 0) : mask(mask), num(num) {}
  bool operator<(const no &q) const { return num < q.num; }
  void print() {
    printf("num=%10d   mask=", num);
    for (int i = 7; i >= 0; i--) printf("%d", (bool)(mask & (1 << i)));
    cout << endl;
  }
};
bool comp(no a, no b) {
  int aa = __builtin_popcount(a.mask), bb = __builtin_popcount(b.mask);
  if (aa != bb) return aa > bb;
  return a.mask > b.mask;
}
set<no> s;
vector<int> v;
vector<no> m;
int show[100];
bool solve(int id, long long k) {
  if ((long long)s.size() < k) return false;
  m.clear();
  for (set<no>::iterator it = s.begin(); it != s.end(); it++)
    m.push_back((*it));
  sort(m.begin(), m.end(), comp);
  for (int i = 0; i < id; i++) show[i] = 0;
  v.clear();
  for (int i = 0; i < k; i++) {
    v.push_back(m[i].num);
    for (int j = 0; j < id; j++) show[j] += !(!(m[i].mask & (1 << j)));
  }
  for (int i = 0; i < id; i++)
    if (show[i] < k / 2) return false;
  for (int i = 0; i < k; i++) {
    if (i != 0) printf(" ");
    printf("%d", v[i]);
  }
  cout << endl;
  return true;
}
int main() {
  int p[9] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
  long long k;
  cin >> k;
  long long val = 2 * k * k;
  s.insert(no(0, 1));
  for (int i = 0; i < 9; i++) {
    set<no> tmp;
    for (set<no>::iterator it = s.begin(); it != s.end(); it++) {
      int pot = p[i];
      while ((*it).num * pot <= val) {
        tmp.insert(no(((*it).mask | (1 << i)), (*it).num * pot));
        pot *= p[i];
      }
    }
    for (set<no>::iterator it = tmp.begin(); it != tmp.end(); it++) {
      s.insert((*it));
    }
    if (solve(i + 1, k)) break;
  }
  return 0;
}