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 <bits/stdc++.h>

#define I inline
#define fi first
#define se second
#define R register
#define LL long long
#define mp make_pair
#define reg register int
#define pii pair<int,int>
#define fo(i, a, b) for(reg i = a; i <= b; i++)
#define fd(i, a, b) for(reg i = a; i >= b; i--)
#define cr const reg&
using namespace std;

I int _max(cr x, cr y) {return x > y ? x : y;}
I int _min(cr x, cr y) {return x < y ? x : y;}
I int read() {
	reg x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
	return x * f;
}
I void ptt(LL x) {if(x >= 10) ptt(x / 10); putchar(x % 10 + '0');}
I void put(LL x) {x < 0 ? putchar('-'), ptt(-x) : ptt(x);}
I void pr1(LL x) {put(x), putchar(' ');}
I void pr2(LL x) {put(x), puts("");}

int main() {
	reg n = read();
	if(n == 3) puts("2 5 63");
	else {
		reg kk = 0;
		while(n >= 8) {
			printf("%d %d %d %d %d %d %d %d ", kk + 2, kk + 3, kk + 4, kk + 6, kk + 8, kk + 9, kk + 10, kk + 12);
			kk += 12, n -= 8;
		} if(n == 1) printf("%d\n", kk + 6);
		else if(n == 2) printf("%d %d\n", kk + 2, kk + 4);
		else if(n == 3) printf("%d %d %d\n", kk + 2, kk + 4, kk + 6);
		else if(n == 4) printf("%d %d %d %d\n", kk + 2, kk + 3, kk + 4, kk + 9);
		else if(n == 5) printf("%d %d %d %d %d\n", kk + 2, kk + 3, kk + 4, kk + 6, kk + 9);
		else if(n == 6) printf("%d %d %d %d %d %d\n", kk + 2, kk + 3, kk + 4, kk + 8, kk + 9, kk + 10);
		else if(n == 7) printf("%d %d %d %d %d %d %d %d\n", kk + 2, kk + 3, kk + 4, kk + 6, kk + 8, kk + 9, kk + 10);
	}
	return 0;
}