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>
#define all(vec) vec.begin(), vec.end()
using namespace std;
using ll = long long;
using P = pair<int, int>;
constexpr ll INF = (1LL << 30) - 1LL;
constexpr ll LINF = (1LL << 60) - 1LL;
constexpr ll MOD = 1e9 + 7;
template <typename T> void chmin(T &a, T b) { a = min(a, b); }
template <typename T> void chmax(T &a, T b) { a = max(a, b); }
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    if (n == 3) {
        cout << 2 << " " << 5 << " " << 63 << endl;
        return 0;
    } else if (n == 4) {
        cout << 2 << " " << 5 << " " << 20 << " " << 63 << endl;
        return 0;
    }
    for (int i = 30000; i >= 2; i -= 2) {
        int k = i / 2;
        if (k < n && k % 3 == 0 && (n - k) % 2 == 0) {
            for (int j = 2; j <= i; j += 2) {
                cout << j << " ";
            }
            n -= k;
            break;
        }
    }
    int t = 3;
    for (int i = 0; i < n; i++) {
        cout << t << " ";
        t += 3;
        if (t % 2 == 0) {
            t += 3;
        }
    }
    cout << endl;
}