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>
#include<stdlib.h>
#include<math.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a,b) ((a)>(0)?(a):-(a))

int func(int s){
  while(s%2==0) s/=2;
  while(s%3==0) s/=3;
  while(s%5==0) s/=5;
  if(s==1) return 0;
  int k=7;
  while(k*k<=s){
    if(s%k==0){
      if(k<=30000) return k;
      return 0;
    }
    k+=2;
  }
  if(s<=30000) return s;
  return 0;
}

void run(void){
  int n;
  //for(n=4;n<=20000;n++){
  scanf("%d",&n);
  if(n==3){
    printf("2 5 63\n");
    return;
    }
  int a[20000];
  int i;
  int k=2;
  int s=0;
  for(i=0;i<n-1;i++){
    while(k%2!=0 && k%3!=0 && k%5!=0) k++;
    a[i]=k;
    s+=k;
    k++;
  }
  if(s%30==0){
    a[i]=30000;
  } else {
    while(!((s+k)%2==0 && (s+k)%3==0 && (s+k)%5==0)) k++;
    int t=func(s+k);
    while(t==0){
      k+=30;
      t=func(s+k);
    }
    a[i]=k;
  }
  for(i=0;i<n;i++){
    printf("%d ",a[i]);
  }
  printf("\n");
  //}
  return;
}

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