Problem: Little Petya has a birthday soon. Due this wonderful event, Petya's friends decided to give him sweets. The total number of Petya's friends equals to n.

Let us remind you the definition of the greatest common divisor: GCD(a1, ..., ak) = d, where d represents such a maximal positive number that each ai (1 ≤ i ≤ k) is evenly divisible by d. At that, we assume that all ai's are greater than zero.

Knowing that Petya is keen on programming, his friends has agreed beforehand that the 1-st friend gives a1 sweets, the 2-nd one gives a2 sweets, ..., the n-th one gives an sweets. At the same time, for any i and j (1 ≤ i, j ≤ n) they want the GCD(ai, aj) not to be equal to 1. However, they also want the following condition to be satisfied: GCD(a1, a2, ..., an) = 1. One more: all the ai should be distinct.

Help the friends to choose the suitable numbers a1, ..., an.

Input

The first line contains an integer n (2 ≤ n ≤ 50).

Output

If there is no answer, print "-1" without quotes. Otherwise print a set of n distinct positive numbers a1, a2, ..., an. Each line must contain one number. Each number must consist of not more than 100 digits, and must not contain any leading zeros. If there are several solutions to that problem, print any of them.

Do not forget, please, that all of the following conditions must be true:

  * For every i and j (1 ≤ i, j ≤ n): GCD(ai, aj) ≠ 1
  * GCD(a1, a2, ..., an) = 1
  * For every i and j (1 ≤ i, j ≤ n, i ≠ j): ai ≠ aj



Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

3


Output

99
55
11115


Input

4


Output

385
360
792
8360

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long normalize(long long x, long long mod) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
long long dx[4] = {0, 0, 1, -1};
long long dy[4] = {1, -1, 0, 0};
string abc = "abcdefghijklmnopqrstuvwxyz";
const double PI = acosl(-1);
const long long mod = 1000 * 1000 * 1000 + 7;
const long long inf = 1e18 + 5;
inline long long add(long long a, long long b, long long p = mod) {
  long long c = a + b;
  if (c >= p) c -= p;
  return c;
}
inline long long sub(long long a, long long b, long long p = mod) {
  long long c = a - b;
  if (c < 0) c += p;
  return c;
}
inline long long mul(long long a, long long b, long long p = mod) {
  return (a * 1ll * b) % p;
}
long long floor1(long long n, long long k) {
  if (n % k == 0 || n >= 0) return n / k;
  return (n / k) - 1;
}
long long ceil1(long long n, long long k) { return floor1(n + k - 1, k); }
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long N = 1e6 + 5;
long long n, p;
long long ans[N];
vector<long long> prim;
void solve() {
  cin >> n;
  if (n == 2) {
    cout << -1;
    return;
  }
  prim.push_back(2);
  prim.push_back(3);
  prim.push_back(5);
  long long pro = 1;
  for (long long i = (1); i < (4); ++i) pro *= prim[(i - 1) % 3];
  long long cnt = 1;
  set<long long> s;
  for (long long i = (1); i < (n + 1); ++i) {
    if ((i - 1) % 3 == 0) ++cnt;
    ans[i] = pro / prim[(i - 1) % 3];
    s.insert(ans[i]);
  }
  while (((long long)s.size()) != n) {
    s.insert(ans[1] * cnt);
    ++cnt;
  }
  for (auto i : s) cout << i << "\n";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}