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: Wrong Answer)

#include<stdio.h>

int gcd(int a,int b){
  int r=a%b;
  while(r>0){
    a=b;b=r;r=a%b;
  }
  return b;
}

void run(void){
  int n;
  scanf("%d",&n);
  int sum=0;
  int cnt=0;
  int i,j;
  for(i=2;cnt<n-2;i++){
    if(i%2==0 || i%3==0){
      printf("%d ",i);
      sum+=i;
      cnt++;
    }
  }
  for(i=5;i<=30000;i++){
    if(i%2==0 || i%3==0) continue;
    for(j=i+2;j<=30000;j++){
      if(j%2==0 || j%3==0) continue;
      if(gcd(sum+i,j)>1 && gcd(sum+j,i)>1){
	printf("%d %d\n",i,j);
	return;
      }
    }
  }
}

int main(void){
  run();
  return 0;
}