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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define sz(a) ((int)a.size())
#define pb push_back

int n;
vector<int> has[3];
vector<int> has3, ans;

int main() {
    // freopen("inp.txt", "r", stdin);

    cin >> n;

    for (int i = 0; i < 15000; i++) {
        int rem = (i + 1) * 2 % 3;
        has[rem].pb((i + 1) * 2);
        if (3 * (2 * i + 1) <= 30000) has3.pb(3 * (2 * i + 1));
    }

    if (n == 3) {
        cout << "2 5 63";
        return 0;
    }

    ans.pb(has[1].back());
    has[1].pop_back();
    ans.pb(has[2].back());
    has[2].pop_back();

    int cur = 0;
    while (sz(ans) < n - 1) {
        if (cur + 1 >= sz(has3)) break;
        ans.pb(has3[cur++]);
        ans.pb(has3[cur++]);
    }


    while (sz(ans) < n - 1) {
        if (has[1].empty()) break;
        ans.pb(has[1].back());
        has[1].pop_back();
        ans.pb(has[2].back());
        has[2].pop_back();
    }

    while (sz(ans) < n) {
        if (has[0].empty()) break;
        ans.pb(has[0].back());
        has[0].pop_back();
    }

    assert(sz(ans) == n);
    for (int i = 0; i < sz(ans); i++) {
        cout << ans[i] << ' ';
    }

    return 0;
}