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 <iostream>
#include <limits.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;

ll gcd(ll a, ll b) {
    if (a < b) {
        ll tmp = a;
        a = b;  b = tmp;
    }
    ll r = a % b;
    while (r != 0) {
        a = b;  b = r;
        r = a % b;
    }
    return b;
}

int main(void){
    int n, ans[20000];
    cin >> n;
    
    int idx = 0;
    int sum = 0;
    for (int i = 2; i <= 30000; i++) {
        if (i % 2 == 0 || i % 3 == 0) {
            ans[idx] = i;  idx++;
            sum += i;
        }
        if (idx == n) break;
    }
    sum -= ans[n];
    
    if (n % 8 == 0 || n % 8 == 7) {
        for (int i = 0; i < n; i++) cout << ans[i] << " ";
        cout << endl;
        return 0;
    }
    
        for (int i = 4; i < 30000; i++) {
            if ((i % 2 != 0 && i % 3 != 0) || ans[n - 2] < i) {
                bool ok = true;
                for (int j = 0; j < n - 1; j++) {
                    //cout << i << " " << sum + i - ans[j] << " " <<  ans[j] << endl;
                    if (gcd(sum + i - ans[j], ans[j]) == 1) {
                        ok = false;  break;
                    }
                }
                //cout << i << " " << sum << " " << i << endl;
                if (gcd(sum, i) == 1) ok = false;
                if (ok) {
                    ans[n - 1] = i;  break;
                }
            }
        }
    
    for (int i = 0; i < n; i++) cout << ans[i] << " ";
    cout << endl;
    
}