Problem:
Score : 600 points

Problem Statement

Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers.

She thinks that a set S = \{a_{1}, a_{2}, ..., a_{N}\} of distinct positive integers is called special if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is not 1.

Nagase wants to find a special set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a special set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.

Constraints

3 \leq N \leq 20000

Input

Input is given from Standard Input in the following format:

N

Output

Output N space-separated integers, denoting the elements of the set S. S must satisfy the following conditions :

The elements must be distinct positive integers not exceeding 30000.

The gcd of all elements of S is 1, i.e. there does not exist an integer d > 1 that divides all elements of S.

S is a special set.

If there are multiple solutions, you may output any of them. The elements of S may be printed in any order. It is guaranteed that at least one solution exist under the given contraints.

Sample Input 1

3

Sample Output 1

2 5 63

\{2, 5, 63\} is special because gcd(2, 5 + 63) = 2, gcd(5, 2 + 63) = 5, gcd(63, 2 + 5) = 7. Also, gcd(2, 5, 63) = 1. Thus, this set satisfies all the criteria.

Note that \{2, 4, 6\} is not a valid solution because gcd(2, 4, 6) = 2 > 1.

Sample Input 2

4

Sample Output 2

2 5 20 63

\{2, 5, 20, 63\} is special because gcd(2, 5 + 20 + 63) = 2, gcd(5, 2 + 20 + 63) = 5, gcd(20, 2 + 5 + 63) = 10, gcd(63, 2 + 5 + 20) = 9. Also, gcd(2, 5, 20, 63) = 1. Thus, this set satisfies all the criteria.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <tuple>

#define int long long
#define double long double

using namespace std;

#define REP(i, a, b) for (int(i) = (a); i < (b); (i)++)
#define rep(i, n) REP(i, 0, n)
#define PER(i, a, b) for (int(i) = (a - 1); i >= (b); (i)--)
#define per(i, n) PER(i, n, 0)
#define each(i, n) for (auto &i : n)
#define clr(a) memset((a), 0, sizeof(a))
#define mclr(a) memset((a), -1, sizeof(a))
#define all(a) (a).begin(), (a).end()
#define dum(val) cerr << #val " = " << val << " ";
#define dump(val) cerr << #val " = " << val << endl;
#define FILL(a, v) fill(a, a + sizeof(a) / sizeof(*a), v)

typedef pair<int, int> P;

const int mod = 1e9 + 7;
const int INF = 1e18;

signed main() {
  int n;
  cin >> n;

  if (n == 3) {
    cout << "2 5 63" << endl;
    return 0;
  }
  if (n == 4) {
    cout << "2 5 20 63" << endl;
    return 0;
  }
  if (n == 6) {
    cout << "2 3 25 55 85 115" << endl;
    return 0;
  }

  if ((n - 5) % 2 == 0) {
    rep(i, 3) { cout << 2 << " "; }
    rep(i, n - 3) {
      if (i == n - 4) {
        cout << 3 << endl;
      } else {
        cout << 3 << " ";
      }
    }
  } else {
    rep(i, 6) { cout << 2 << " "; }
    rep(i, n - 6) {
      if (i == n - 7) {
        cout << 3 << endl;
      } else {
        cout << 3 << " ";
      }
    }
  }
}

/*
  2 4 8 16

  a = odd & gcd(a + 2, 3) != 1

*/