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 <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

int n;
vector <int> a;

int main() {

    scanf("%d", &n);

    if (n == 3) {
        printf("2 3 25\n");
    } else if (n == 4) {
        printf("2 3 4 9\n");
    } else {
        vector <int> a;
        a.push_back(2);
        a.push_back(3);
        a.push_back(4);
        a.push_back(29997);
        for (int i = 9; i + 6 <= 30000 && a.size() + 2 <= n; i += 12) {
            a.push_back(i);
            a.push_back(i + 6);
        }
        for (int i = 6; i <= 30000 && a.size() < n; i += 6) {
            a.push_back(i);
        }
        if (a.size() != n) {
            puts("-1");
            exit(0);
        }
        printf("%d", a[0]);
        for (int i = 1; i < n; i++) {
            printf(" %d", a[i]);
        }
        puts("");
    }

    return 0;

}