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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int gcd(int x, int y)
{
    int r;
    r = x % y;
    while (r != 0)
    {
        x = y;
        y = r;
        r = x % y;
    }
    return y;
}
int getrand(int min, int max)
{
    return min + (int)(rand()*(max - min + 1) / (1 + RAND_MAX));
}
int main()
{
    int n;
    scanf("%d", &n);
    int ans[20000];
    int i;
    int x;
    int sum;
    int j;
    ans[0] = 2;
    ans[1] = 5;
    ans[2] = 63;
    ans[3] = 20;
    for (;;)
    {
        srand((unsigned int)time(NULL));
        sum = 90;
        x = 0;
        for (i = 4; i < n; i++)
        {
            x = 0;
            ans[i] = getrand(1, 30000);
            for (j = 0; j < i; j++)
            {
                if (ans[i] == ans[j])
                {
                    i -= 1;
                    x++;
                    break;
                }
            }
            if(x==0)
                sum += ans[i];
        }
        x = 0;
        for (i = 0; i < n; i++)
        {
            if (gcd(ans[i], sum - ans[i]) == 1)
            {
                x++;
                break;
            }
        }
        if (x == 0)
        {
            for (i = 0; i < n - 1; i++)
                printf("%d ", ans[i]);
            printf("%d\n", ans[n - 1]);
            return 0;
        }
    }
}