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<cstdio>
const int MAXN=20005,MAXV=30001;

int ans[MAXN];
bool used[MAXV];
int gcd(int a,int b)
{return b==0?a:gcd(b,a%b);}

int main()
{
    int n,sum=0,i,j;
    scanf("%d",&n);
    for(i=1,j=1;j<n;i++)//添加n-1个2、3的倍数
        if(i%2==0||i%3==0)
        {
            ans[j++]=i;
            used[i]=true;
            sum+=i;
        }
    int r2=sum%2,r3=sum%3;
    for(i=1;j<=n;i++)//枚举最后一个数
        if(!used[i]&&(i+r2)%2==0&&(i+r3)%3==0&&gcd(i,sum)!=1)
            ans[j++]=i;
    for(int i=1;i<n;i++)
        printf("%d ",ans[i]);
    printf("%d\n",ans[n]);

    return 0;
}